WHOEVER ANSWERS IS THE BRAINLIEST!!! PLS HELP!!

Guiseppe's buys supplies to make pizzas at a cost of $4.02. Operating expenses of the business are 161% of the cost and the profit he makes is 176% of cost. The regular selling price of each pizza is $7.33.

Let's denote the cost of supplies as C.

Operating expenses:

The operating expenses of the business are 161% of the cost. Therefore, the operating expenses can be calculated as:

Operating Expenses = 1.61 * C

Profit:

The profit made by Guiseppe is 176% of the cost. Therefore, the profit can be calculated as:

Profit = 1.76 * C

Total cost:

The total cost includes the cost of supplies and the operating expenses:

Total Cost = C + Operating Expenses = C + 1.61 * C = 2.61 * C

Regular selling price:

The regular selling price is the sum of the total cost and the profit:

Regular Selling Price = Total Cost + Profit = 2.61 * C + 1.76 * C = 4.37 * C

Given that the cost of supplies is $4.02, we can substitute this value into the equation:

Regular Selling Price = 4.37 * 4.02 = $17.5674

Rounding the final answer to the nearest cent, the regular selling price of each pizza is approximately $7.33.

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Answer:

To solve this problem, we need to use some principles of physics, specifically Newton's second law (F=ma) and the equations of motion. Here are the steps:

1. Calculate the acceleration (a)

We can use the equation of motion to find the acceleration:

v_f^2 = v_i^2 + 2a*d

where:

v_f = final velocity = 6.80 m/s

v_i = initial velocity = 3.00 m/s

d = distance = 2/3 of the length of the fish = 2/3 * 1.50 m = 1.00 m

a = acceleration (which we are trying to find)

Rearranging the equation to solve for a gives us:

a = (v_f^2 - v_i^2) / (2*d)

2. Calculate the magnitude of the force F

Once we have the acceleration, we can use Newton's second law (F=ma) to calculate the force. The net force acting on the fish as it jumps out of the water is the difference between the upward force F exerted by the tail fin and the downward force due to gravity (mg). The net force is also equal to the product of the mass of the fish and its acceleration (ma). Therefore, we have:

F - mg = ma

Rearranging this equation to solve for F gives us:

F = ma + mg

Now let's plug in the numbers and do the calculations.

First, let's find the acceleration:

a = (v_f^2 - v_i^2) / (2*d)

a = (6.80 m/s)^2 - (3.00 m/s)^2) / (2*1.00 m)

a = (46.24 m^2/s^2 - 9.00 m^2/s^2) / 2 m

a = 37.24 m^2/s^2 / 2 m

a = 18.62 m/s^2

The salmon's acceleration is 18.62 m/s^2 upward.

Next, let's find the force F. We know the mass of the fish is 52.0 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. So,

F = ma + mg

F = (52.0 kg)(18.62 m/s^2) + (52.0 kg)(9.8 m/s^2)

F = 969.24 N + 509.6 N

F = 1478.84 N

So, the magnitude of the force F exerted by the salmon's tail fin during this interval is approximately 1479 N.

The magnitudes of the tensions are T1 = mg sin(θ) and T2 = 2mg cos(θ).

In order to find the tensions T1 and T2 in the given system, let's analyze the forces acting on the two blocks. We assume that the incline makes an angle θ with the horizontal.

For the block of mass m, the forces acting on it are its weight mg acting vertically downwards and the tension T1 acting along the incline. The weight can be split into two components: mg sin(θ) perpendicular to the incline and mg cos(θ) parallel to the incline. Since the block is in equilibrium, the sum of the forces along the incline must be zero. Therefore, T1 = mg sin(θ).

For the block of mass 2m, the forces acting on it are its weight 2mg vertically downwards, the tension T2 acting vertically upwards, and the tension T1 acting along the incline. The sum of the forces along the incline for this block is also zero. Therefore, T1 = 2mg sin(θ) and T2 = 2mg cos(θ).

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a.  the tension in the rope is  91.5 N.

b.   the moment of inertia of the wheel is  0.1008 kg⋅m².

c.  the angular speed of the wheel 2.30 s after it begins rotating is  38.34 rad/s.

(a)

The tension in the rope can be found by considering the forces acting on the object.

ma = mg*sin(θ) - T

(14.0 kg)(2.00 m/s²)

= (14.0 kg)(9.8 m/s²)*sin(37°) - T

T = (14.0 kg)(9.8 m/s²)*sin(37°) - (14.0 kg)(2.00 m/s²)

T =  91.5 N

(b)

The moment of inertia of a wheel:

I = (1/2)MR²

I = (1/2)(14.0 kg)(0.12 m)²

I = 0.1008 kg⋅m²

(c)

The angular acceleration of the wheel:

α = a/R

α = angular acceleration,

a = linear acceleration of the object,

R =  radius of the wheel.

α = (2.00 m/s²)/(0.12 m)

α = 16.67 rad/s²

The angular speed (ω) of the wheel after time t is :

ω = ω₀ + αt

ω = 0 + (16.67 rad/s²)(2.30 s)

ω = 38.34 rad/s

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(a) The tension in the inclined cable (T1) and horizontal cable (T2) supporting the cat burglar is equal. The tension in the vertical cable (T3) is 524 N.

(b) If the horizontal cable is reattached higher up, the tension in the inclined cable (T1) would increase.

(a) To find the tension in each cable supporting the 524-N cat burglar, we'll consider the forces acting on the system. Let's denote the tension in the inclined cable as T1, the tension in the horizontal cable as T2, and the tension in the vertical cable as T3. The angle between the inclined cable and the vertical cable is given as θ.

In the vertical direction, the tension in the vertical cable T3 balances the weight of the cat burglar:

T3 - 524 N = 0

T3 = 524 N

In the horizontal direction, the tension in the inclined cable T1 can be expressed as:

T1 * cos(θ) = T2

Now, we need to determine the value of θ to calculate T1 and T2. Let's assume that θ is the given angle of θ = 0.

Substituting the angle and rearranging the equation, we have:

T1 = T2 / cos(θ)

T1 = T2 / cos(0)

T1 = T2 / 1

T1 = T2

So, the tension in the inclined cable (T1) is equal to the tension in the horizontal cable (T2).

Therefore, the tension in each cable is as follows:

T1 (inclined cable) = T2 (horizontal cable)

T1 = T2

T3 (vertical cable) = 524 N

(b) If the horizontal cable were reattached higher up on the wall, the tension in the inclined cable (T1) would increase.

The correct answer is option A.  

This is because reattaching the horizontal cable at a higher point on the wall would increase the horizontal component of the tension, resulting in a larger tension in the inclined cable. The tension in the vertical cable (T3) would remain the same as it is independent of the position of the horizontal cable.

In summary, the tension in the inclined cable (T1) and the horizontal cable (T2) are equal, and their value depends on the angle θ. The tension in the vertical cable (T3) is 524 N. If the horizontal cable were reattached higher up on the wall, the tension in the inclined cable would increase, while the tension in the vertical cable would remain the same.

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The cardinality of R₁ \ (R₁ intersection A) is 4. The given polynomial (2x² + y² + 22 - 8x + 2y - 2xy + 2xz-16z + 35) cannot be solved for x, y, and z due to insufficient equations.

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(a) The speed at which the ball was launched is approximately 10.91 m/s.

(b) The ball clears the wall by approximately 1.50 m vertically.

(c) The horizontal distance from the wall to the point on the roof where the ball lands is approximately 24.02 m.

To solve this problem, we'll analyze the motion of the ball in two dimensions: horizontal and vertical.

(a) First, let's calculate the initial speed at which the ball was launched. We can use the time of flight and the horizontal distance traveled to find the initial horizontal velocity (Vx) of the ball.

The horizontal distance traveled by the ball (d) is given as 24.0 m, and the time of flight (t) is given as 2.20 s.

Using the equation for horizontal distance:

d = Vx * t

Rearranging the equation, we can solve for Vx:

Vx = d / t

Plugging in the known values:

Vx = 24.0 m / 2.20 s

Simplifying the equation, we find:

Vx ≈ 10.91 m/s

The initial horizontal velocity of the ball is approximately 10.91 m/s.

(b) Next, let's find the vertical distance by which the ball clears the wall. We can use the time of flight and the vertical motion of the ball to calculate this.

The vertical distance traveled by the ball is the difference between the height of the wall (h) and the height of the playground (hb).

Δy = h - hb

Plugging in the known values:

Δy = 7.40 m - 5.90 m

Simplifying the equation, we find:

Δy = 1.50 m

The ball clears the wall by approximately 1.50 m vertically.

(c) Finally, let's determine the horizontal distance from the wall to the point on the roof where the ball lands.

Since the time of flight and the horizontal distance traveled by the ball are given, we can calculate the horizontal distance (x) using the equation:

x = Vx * t

Plugging in the known values:

x = 10.91 m/s * 2.20 s

Simplifying the equation, we find:

x ≈ 24.02 m

The ball lands approximately 24.02 m horizontally from the wall on the roof.

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a) The speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

b) We cannot calculate the work done by the friction force.

c) The net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

(a) The work done on the 6.00 kg block by gravity can be calculated using the formula:

Work_gravity = force_gravity * displacement * cos(theta),

where force_gravity is the weight of the block, displacement is the downward displacement of the block, and theta is the angle between the force and displacement vectors (which is 0 degrees in this case).

The weight of the block is given by:

force_gravity = mass * acceleration_due_to_gravity = 6.00 kg * 9.8 m/s^2 = 58.8 N.

Plugging in the values, we get:

Work_gravity = 58.8 N * 0.800 m * cos(0) = 47.04 J.

The work done on the 6.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(180) = -29.6 J.

The negative sign indicates that the tension is in the opposite direction of the displacement.

Using the work-energy theorem, we can find the speed of the 6.00 kg block after descending 0.800 m:

Work_net = change_in_kinetic_energy.

Since the block starts from rest, its initial kinetic energy is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * mass * velocity^2.

Solving for velocity, we get:

velocity = sqrt(2 * Work_net / mass).

The net work done on the block is the sum of the work done by gravity and the tension:

Work_net = Work_gravity + Work_tension = 47.04 J - 29.6 J = 17.44 J.

Plugging in the values, we get:

velocity = sqrt(2 * 17.44 J / 6.00 kg) = 2.07 m/s.

Therefore, the speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

(b) The total work done on the 8.00 kg block during the 0.800 m displacement can be calculated using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the 8.00 kg block is not moving vertically, its initial and final kinetic energies are zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 0.

The work done on the 8.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(0) = 29.6 J.

The work done on the 8.00 kg block by the friction force can be calculated using the formula:

Work_friction = force_friction * displacement * cos(theta),

where force_friction is the frictional force on the block. However, the problem statement does not provide the value of the friction force. Therefore, we cannot calculate the work done by the friction force.

(c) The net work done on the system of two blocks during the 0.800 m displacement of the 6.00 kg block can be found using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the system starts from rest, the initial kinetic energy of the system is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * (6.00 kg + 8.00 kg) * velocity^2.

Simplifying, we get:

Work_net = 1/2 * 14.00 kg * velocity^2.

Using the value of velocity calculated in part (a), we get:

Work_net = 1/2 * 14.00 kg * (2.07 m/s)^2 = 29.13 J.

The work done on the system of two blocks by gravity is the sum of the work done on the individual blocks by gravity:

Work_gravity_system = Work_gravity_6kg + Work_gravity_8kg = 47.04 J + 0 J = 47.04 J.

The work done on the system of two blocks by the tension in the rope is the sum of the work done on the individual blocks by the tension:

Work_tension_system = Work_tension_6kg + Work_tension_8kg = -29.6 J + 29.6 J = 0 J.

Therefore, the net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

Note: The calculations for part (b) and (c) were based on the given information, but the value of the friction force was not provided in the problem statement.

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The velocity of the third piece is v₃ = -12500 kg·m/s / m₃

The law of conservation of momentum states that the total momentum before the explosion is equal to the total momentum after the explosion.

velocity of the third piece =  v₃.

The total initial momentum before the explosion = 0

The total final momentum after the explosion= 0

Initial momentum = 0 kg·m/s (since the bomb is at rest)

Final momentum = m₁v₁ + m₂v₂ + m₃v₃

m₁ = mass of the first piece = 150 kg

v₁ = velocity of the first piece = 150 m/s (to the east)

m₂ = mass of the second piece = 100 kg

v₂ = velocity of the second piece = 200 m/s (south 60° west)

m₃ = mass of the third piece = unknown

v₃ = velocity of the third piece = unknown

0 = (150 kg)(150 m/s) + (100 kg)(200 m/s)(cos(60°)) + (m₃)(v₃)

final momentum = 0 and hence  v₃ is found as :

0 = 22500 kg·m/s - 10000 kg·m/s + (m₃)(v₃)

-12500 kg·m/s = (m₃)(v₃)

v₃ = -12500 kg·m/s / m₃

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The six digit grid coordinates for the windtee  should be 100049.

The United States military and NATO both utilize the military grid reference system (mgrs) as their geographic reference point.

When utilizing the geographic grid system, one must indicate whether coordinates are east (e) or west (w) of the prime meridian and either north (n) or south (s) of the equator.

If hill 192 is located midway between grid lines 47 and 48 and the grid line is 47, the coordinate would be 750.

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The velocity of the third piece is (81.25 m/s, -43.3 m/s).

To determine the velocity of the third piece, we can use the principle of conservation of momentum.

Given:

Mass of the first piece (m1) = 150 kg

Velocity of the first piece (v1) = 150 m/s (to the East)

Mass of the second piece (m2) = 100 kg

Velocity of the second piece (v2) = 200 m/s at a direction of south 60° West

Let's break down the velocities into their respective horizontal (x) and vertical (y) components.

For the first piece:

v1x = 150 m/s (since it's moving to the East)

v1y = 0 m/s (no vertical component)

For the second piece:

v2x = 200 m/s * cos(60°) = 200 m/s * 0.5 = 100 m/s (horizontal component)

v2y = -200 m/s * sin(60°) = -200 m/s * 0.866 = -173.2 m/s (vertical component, negative since it's moving downward)

Now, let's calculate the momentum of the first and second pieces:

The momentum of the first piece (p1) = m1 * v1

= 150 kg * 150 m/s

= 22,500 kg·m/s

The momentum of the second piece (p2) = m2 * v2

= 100 kg * (100 m/s, -173.2 m/s)

= (10,000 kg·m/s, -17,320 kg·m/s)

To find the total momentum after the explosion, we can add the momenta of the individual pieces:

Total momentum after the explosion = p1 + p2

= (22,500 kg·m/s, 0 kg·m/s) + (10,000 kg·m/s, -17,320 kg·m/s)

= (32,500 kg·m/s, -17,320 kg·m/s)

The total momentum after the explosion should also be equal to the momentum of the third piece:

The momentum of the third piece (p3) = m3 * v3

Given:

Mass of the third piece (m3) = 400 kg (calculated from the given mass of the bomb)

Let's assume the velocity of the third piece is (v3x, v3y).

Therefore, we have the equation:

(32,500 kg·m/s, -17,320 kg·m/s) = 400 kg * (v3x, v3y)

By equating the x and y components separately, we can solve for the velocity components of the third piece:

32,500 kg·m/s = 400 kg * v3x

-17,320 kg·m/s = 400 kg * v3y

Solving these equations, we find:

v3x = 81.25 m/s

v3y = -43.3 m/s

Therefore, the velocity of the third piece is approximately (81.25 m/s, -43.3 m/s).

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(a) The acceleration of the system is 8.5 m/s².

(b) The tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.

(c) The force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.

To solve this problem, we can use Newton's second law of motion (F = ma) and consider the forces acting on each block individually.

(a) Determine the acceleration given this system:

To find the acceleration (a) of the system, we can use the net force acting on the 4.0 kg block (m3). The only force acting on m3 is the applied force (F = 34 N).

F = m3 * a

34 N = 4.0 kg * a

Solving for a, we find:

a = 34 N / 4.0 kg

a = 8.5 m/s²

Therefore, the acceleration of the system is 8.5 m/s².

(b) Determine the tension in the cord connecting the 4.0-kg and the 1.0-kg blocks:

To find the tension in the cord (T), we can consider the forces acting on the 1.0 kg block (m1).

T - F = m1 * a

T - 34 N = 1.0 kg * 8.5 m/s²

T - 34 N = 8.5 N

T = 42.5 N

Therefore, the tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.

(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block:

To find the force exerted by the 1.0 kg block (m1) on the 2.0 kg block (m2), we can consider the forces acting on the 2.0 kg block.

F - T = m2 * a

F - 42.5 N = 2.0 kg * 8.5 m/s²

F - 42.5 N = 17 N

F = 59.5 N

Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.

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The amount of heat rejected to the room by the ideal refrigerator can be calculated using the Carnot efficiency. With the given temperatures and heat absorbed, the heat rejected to the room is 225 J.

To calculate the amount of heat rejected to the room by the ideal refrigerator, we can use the Carnot efficiency, which is given by the formula:

Efficiency = 1 - ([tex]T_c_o_l_d[/tex] / [tex]T_h_o_t[/tex])

where[tex]T_c_o_l_d[/tex]is the temperature of the cold reservoir (freezer compartment) and [tex]T_h_o_t[/tex] is the temperature of the hot reservoir (room temperature).

Given:

[tex]T_c_o_l_d[/tex] = -9 °C (converted to Kelvin: 264 K)

[tex]T_h_o_t[/tex]= 25 °C (converted to Kelvin: 298 K)

Heat absorbed from the freezer compartment ([tex]Q_c_o_l_d[/tex] = 120 J

First, we calculate the Carnot efficiency:

Efficiency = 1 - (264 K / 298 K)

Efficiency ≈ 0.1134

The Carnot efficiency represents the ratio of heat transferred from the cold reservoir to the work done by the refrigerator. Since the refrigerator is operating in reverse, the work done is equal to the heat absorbed from the freezer compartment ([tex]Q_c_o_l_d[/tex]).

[tex]Q_c_o_l_d[/tex] = 120 J

Now, we can calculate the heat rejected to the room ([tex]Q_h_o_t[/tex]) using the equation:

[tex]Q_h_o_t[/tex] = Efficiency * [tex]Q_c_o_l_d[/tex]

[tex]Q_h_o_t[/tex] ≈ 0.1134 * 120 J

[tex]Q_h_o_t[/tex] ≈ 13.61 J

Therefore, the amount of heat rejected to the room by the ideal refrigerator is approximately 13.61 J.

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The six digit grid coordinates for the windtee  should be 3.

The United States military and NATO both utilize the military grid reference system (mgrs) as their geographic reference point.

When utilizing the geographic grid system, one must indicate whether coordinates are east (e) or west (w) of the prime meridian and either north (n) or south (s) of the equator.

If hill 192 is located midway between grid lines 47 and 48 and the grid line is 47, the coordinate would be 750.

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The six digit grid coordinates for the windtee is determined as 100049.

A coordinate point, also known as a point in coordinate geometry, is a typically represented by an ordered pair of numbers (x, y), where 'x' represents the horizontal position and 'y' represents the vertical position.

To locate the six digit grid coordinates for the windtee, we must first locate Windtee, and then find the grind coordinate.

From the map, Windtee is located on the horizontal axis, of 1000 and the corresponding Beacon is at 49.

So the six digit grid coordinates = 100049.

Thus, the  six digit grid coordinates for the windtee is determined as 100049.

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The graphic organizer suggests that the job of an astronaut is complex, demanding, and involves flight.

This conclusion can be drawn by examining the nature of the requirements listed in the graphic organizer. Firstly, the requirements listed in the organizer are numerous and encompass various aspects. They include educational qualifications, such as having a bachelor's degree in a relevant field, as well as specific experience, like piloting an aircraft.

These requirements highlight the complexity of the job and indicate that astronauts need to possess a diverse set of skills and knowledge. Additionally, the requirements for physical fitness and health demonstrate the demanding nature of the job.

Astronauts are expected to undergo rigorous physical training to ensure they can handle the physical stresses associated with space travel and the conditions they will encounter in space. This indicates that the job can be physically exhausting and requires individuals to be in excellent health.

Lastly, the inclusion of flight-related requirements, such as the need to pass a long-duration spaceflight physical and participate in aircraft flights, implies that the job of an astronaut involves actual flight experiences. This indicates that astronauts are directly involved in piloting spacecraft and are expected to have practical experience in flying.

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The total volume of the piece of wood is 1.33[tex]cm^3[/tex].

To calculate the total volume of the piece of wood, we can use the principle of displacement.

1. First, we need to find the difference in volume between the two water levels. The initial volume is 17.7 [tex]cm^3[/tex], and the final volume is 18.5 cm^3. The difference is 18.5 [tex]cm^3[/tex] - 17.7 [tex]cm^3[/tex] = 0.8 [tex]cm^3[/tex].

2. Now, we need to find the volume of water displaced by the piece of wood. Since the relative density of the wood is 0.60, it means that the wood is 0.60 times denser than water.

3. The volume of water displaced by the wood is equal to the difference in volume divided by the relative density of the wood. So, the volume of water displaced is 0.8 cm^3 / 0.60 = 1.33 [tex]cm^3[/tex].

4. Finally, the total volume of the piece of wood is equal to the volume of water displaced. Therefore, the total volume of the piece of wood is 1.33 [tex]cm^3[/tex].

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The acceleration of the two blocks is[tex]2.14 m/s^{2[/tex]} and the distance does block M2 move in 2.00 s is 4.27 m.

Now we need to find the acceleration of the two blocks and the distance does block M2 move in 2.00 s.

We know that: mass of M1, m1 = 5.00 kg mass of M2, m2 = 19.0 kgθ = 25.0°Taking upward direction as positive for block M1 and downwards as positive for block M2.

Therefore, we can write the following equation of motion for the two blocks:

For M2: m2g - T = m2a ...(1)

For M1: T - m1g = m1a ...(2)

We can see from the figure that M2 is on an inclined plane making an angle θ with the horizontal.

We can resolve the weight of M2 into two components:

Perpendicular to the plane = m2gcosθParallel to the plane = m2gsinθ

The component parallel to the plane will tend to make the block move downwards.

Therefore, the effective weight will be:

mg = m2gsinθ ...(3)

From equation (1) we can write:

T = m2g - m2a ...(4)

Substituting equation (4) in equation (2), we get:

m2g - m2a - m1g = m1a ...(5)

On solving equation (5), we get the acceleration as:

a = g(m2sinθ - m1) / (m1 + m2)

On substituting the given values, we get:

[tex]a = 2.14 m/s^{2}[/tex]

The distance moved by M2 in 2 seconds can be found out using the formula:[tex]s = ut + \frac{1}{2} at^{2}[/tex]

Here, initial velocity, u = 0m/s Time, t = 2s Acceleration, [tex]a = 2.14 m/s^{2}[/tex]

On substituting these values, we get the distance travelled by M2 as: s = 4.27 m

Therefore, the acceleration of the two blocks is [tex]2.14 m/s^{2}[/tex]. And the distance does block M2 move in 2.00 s is 4.27 m.

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Answer:

C

Explanation:

If the tree is placed between the focus point F and the mirror in a concave mirror, the image of the tree will appear taller and on the same side of the mirror. Therefore, the correct answer is C. The image appears taller and on the same side of the mirror.

Answer:

Approximately [tex]1.89\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

Let [tex]m_{A}[/tex] and [tex]m_{B}[/tex] denote the mass of the two vehicles. Let [tex]u_{A}[/tex] and [tex]u_{B}[/tex] denote the velocity before the collision. Let [tex]v_{A}[/tex] and [tex]v_{B}[/tex] denote the velocity after the collision.

Since the collision is elastic, both momentum and kinetic energy should be conserved.

For momentum to conserve:

[tex]m_{A} \, v_{A} + m_{B} \, v_{B} = m_{A}\, u_{A} + m_{B}\, u_{B}[/tex].

For kinetic energy to conserve:

[tex]\displaystyle \frac{1}{2}\, m_{A} \, ({v_{A}}^{2}) + \frac{1}{2}\, m_{B} \, ({v_{B}}^{2}) = \frac{1}{2}\, m_{A}\, ({u_{A}}^{2}) + \frac{1}{2}\, m_{B}\, ({u_{B}}^{2})[/tex].

Simplify to obtain:

[tex]\displaystyle m_{A} \, ({v_{A}}^{2}) + m_{B} \, ({v_{B}}^{2}) = m_{A}\, ({u_{A}}^{2}) + m_{B}\, ({u_{B}}^{2})[/tex].

It is given that [tex]m_{A} = 282\; {\rm kg}[/tex], [tex]m_{B} = 210\; {\rm kg}[/tex], [tex]u_{A} = 2.82\; {\rm m\cdot s^{-1}}[/tex], and [tex]u_{B} = 1.72\; {\rm m\cdot s^{-1}}[/tex]. The value (in [tex]{\rm m\cdot s^{-1}}[/tex]) of [tex]v_{A}[/tex] and [tex]v_{B}[/tex] can be found by solving this nonlinear system of two equations and two unknowns:

[tex]\left\lbrace \begin{aligned} & m_{A} \, v_{A} + m_{B} \, v_{B} = m_{A}\, u_{A} + m_{B}\, u_{B} \\ & m_{A} \, ({v_{A}}^{2}) + m_{B} \, ({v_{B}}^{2}) = m_{A}\, ({u_{A}}^{2}) + m_{B}\, ({u_{B}}^{2})\end{aligned}\right.[/tex].

[tex]\left\lbrace \begin{aligned} & 282 \, v_{A} + 210 \, v_{B} = 282\, (2.82) + 210\, (1.72) \\ & 282 \, ({v_{A}}^{2}) + 210 \, ({v_{B}}^{2}) = 282\, ({2.82}^{2}) + 210\, ({1.72}^{2})\end{aligned}\right.[/tex].

Solving this system gives two possible sets of solutions:

However, the second set of solutions is invalid since it suggests that the velocity of the two vehicles stayed unchanged after the collision. Hence, only the first set of solutions ([tex]v_{A} &\approx 1.89\; {\rm m\cdot s^{-1}}[/tex], [tex]v_{B} &\approx 2.98\; {\rm m\cdot s^{-1}}[/tex]) is valid.

Therefore, the velocity of vehicle [tex]A[/tex] would be approximately [tex]1.89\; {\rm m\cdot s^{-1}}[/tex] after the collision.

(a) The tension in the string is determined as 19.6 N.

(b) The acceleration of each object is 5.3 m/s².

(c) The distance each object will move in the first second if it started from rest is 2.65 m.

(a) The tension in the string is the resultant weight of the masses and magnitude is calculated as follows;

T = ( 5.7 kg - 3.7 kg ) x 9.8 m/s²

T = 19.6 N

(b) The acceleration of each object is calculated as follows;

a = T / m

where;

a = 19.6 N / 3.7 kg

a = 5.3 m/s²

(c) The distance each object will move in the first second if it started from rest is calculated as;

s = ut + ¹/₂at²

where;

s = 0 + ¹/₂(5.3)(1²)

s = 2.65 m

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When white light reflects off of a green surface, only the green wavelengths of light are reflected (option d).

1. White light is a combination of all visible wavelengths of light, including red, orange, yellow, green, blue, indigo, and violet.

2. When white light hits a green surface, the surface absorbs some wavelengths of light and reflects others.

3. The color we perceive as "green" is the result of the green wavelengths of light being reflected by the surface.

4. In this case, the green surface absorbs all the wavelengths of light except for the green wavelengths, which are reflected back.

5. As a result, our eyes detect the reflected green light and interpret it as the color green.

6. This phenomenon occurs because the green surface selectively absorbs and reflects different wavelengths of light based on its molecular structure and the interactions between light and matter.

7. The absorption and reflection of specific wavelengths of light give objects their perceived color.

8. Therefore, when white light reflects off of a green surface, only the green wavelengths of light are reflected, while the other wavelengths are absorbed by the surface.

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It takes about 4.85 seconds for the car to accelerate from a speed of 25 mi/h to a speed of 32 mi/h.

The given information includes the acceleration rate of a certain car which is 0.65 m/s², and the initial speed of the car which is 25 miles per hour. The question is asking about the time taken by the car to accelerate from the initial speed of 25 miles per hour to a speed of 32 miles per hour. This is a simple problem in kinematics that can be solved by using the formula of acceleration. Here’s how:
First, convert the initial and final speeds of the car into meters per second.
Given that:
Initial speed of the car, u = 25 miles/hour
Final speed of the car, v = 32 miles/hour
To convert miles/hour to meters/second, multiply it by 0.447:
u = 25 miles/hour × 0.447 = 11.175 meters/second
v = 32 miles/hour × 0.447 = 14.324 meters/second
Now, let’s use the formula of acceleration:
v = u + at
Where,
v = final speed = 14.324 m/s
u = initial speed = 11.175 m/s
a = acceleration = 0.65 m/s²
t = time taken
Substitute the given values in the formula:
14.324 = 11.175 + (0.65)t
Solve for t:
t = (14.324 - 11.175) / 0.65
t = 4.85 seconds
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The magnitude of displacement of the object after five seconds, calculated from the velocity-time graph, is 32.5 m. The correct answer is option E.

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In this case, the demand (D1) moves to the left (D2), this also happens with supply (S1) leading to (S2), moreover, the intersections between these lines represent E1, E2, and E3.

Due to an increase in salary, it is expected the demand for dinners increase, which means this line would move to the left. This occurs as a higher wage for everyone implies people are more willing to pay for dinner than before.

This change would also mean restaurants are likely to provide more quantity, which increases the supply, and therefore in this process the equilibrium changes.

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The function of the power supply is to provide electrical energy to the device or system that needs it. The power supply converts the incoming voltage from the power source into a form that is usable by the device, such as DC voltage.

The readout is a device or component that displays data or information to the user. The readout could be a simple LED display or a complex graphical display.

A peripheral is a device or component that connects to a computer or other electronic device to provide additional functionality. Examples of peripherals include printers, scanners, and external hard drives.

A microcomputer is a type of computer that is designed to fit on a single microchip. Microcomputers are found in a wide range of devices, including smart phones, tablets, and embedded systems.

A transducer is a device that converts one form of energy to another. In electronics, transducers are commonly used to convert electrical energy into mechanical energy, or vice versa.

The processor is the central component of a computer or electronic device. The processor is responsible for executing instructions and controlling the other components of the system. The performance and capabilities of a device are largely determined by the speed and power of the processor.

The force that acts on a projectile in the horizontal direction is Gravitational force.

A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.

Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity. Horizontal forces are not required to keep a projectile moving horizontally. Hence, The only force acting upon a projectile is gravity.

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A 0.5 kg soccer ball is kicked at 10 m/s. A goalie catches the ball and brings it to rest in 0.25 seconds then the force exerted on the ball by the goalie is 20N. Option C is correct.

Here, we must determine the change in momentum of the soccer ball. The momentum of an object is stated by the product of its mass and velocity. The mass of the soccer ball is 0.5 kg, and its initial velocity is 10 m/s. Therefore, the ball is conducted to be constant, and its final velocity is 0 m/s.

The change in momentum is computed by reducing the final momentum from the initial momentum. In this concern, the initial momentum is 0.5 kg × 10 m/s = 5 kg·m/s, and the final momentum is 0.5 kg × 0 m/s = 0 kg·m/s. Now, the change in momentum is 5 kg·m/s - 0 kg·m/s = 5 kg·m/s.

Next, we separate the change in momentum by the time taken to bring up the ball to rest, which is 0.25 seconds. Thus, the goalie's force exerted on the ball is 5 kg·m/s / 0.25 s = 20 N.

Therefore, the correct answer is C. 20 N.

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The correct Option is C. The force exerted on the ball by the goalie is 20 N.

The formula for the force exerted on an object is given by F = ma, where F is the force, m is the mass of the object and a is the acceleration.

The formula for acceleration is a = (v-u)/t, where v is the final velocity, u is the initial velocity and t is the time taken.

The acceleration is negative if the object is brought to rest.

So, for the given problem, the initial velocity of the soccer ball is 10 m/s and the final velocity is 0.

The time taken to bring it to rest is 0.25 s.

Therefore, the acceleration is given by:a = [tex](0 - 10)/0.25 = - 40 m/s^{2}[/tex]

Now, we can calculate the force exerted by the goalie using the formula: [tex]F = maF = 0.5 kg $\times$ (- 40 m/s^{2} ) = - 20 N[/tex]

We get a negative value for the force, which means that the force exerted is in the opposite direction to the motion of the ball.

However, the magnitude of the force is given by |-20 N| = 20 N.

So, the answer is option (C) 20 N.

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The height of the building is approximately 32.34 meters.

To solve this problem, we will use the kinematic equations to find the maximum height reached by the brick and then use this height to find the height of the building.

We can start by breaking the initial velocity of the brick into its horizontal and vertical components as follows:

v₀x = v₀cos(θ) = 17cos(25°) ≈ 15.84 m/s

v₀y = v₀sin(θ) = 17sin(25°) ≈ 7.23 m/s

where θ is the angle of the initial velocity to the horizontal.

Next, we can use the following kinematic equation to find the maximum height reached by the brick:

y = y₀ + v₀yt - 1/2gt²

where y₀ is the initial height (height of the building), t is the time of flight, and g is the acceleration due to gravity (9.81 m/s²).

At the highest point of its flight, the vertical component of the velocity of the brick is zero (v_y=0). We can use this fact to find the time taken to reach maximum height:

v_y = v₀y - gt

0 = v₀y - gt_max

t_max = v₀y / g ≈ 0.738 s

We can then substitute this value of t_max into the expression for y to obtain the maximum height:

y_max = y₀ + v₀y t_max - 1/2 g t_max²

where we set y = y_max and t = t_max.

Next, we can use the total flight time of the brick (3.1 s) to find the initial height of the building:

3.1 = t_max + t_down

where t_down is the time taken by the brick to fall from the maximum height to the ground. Since the brick falls down for the same time as it takes to go up, we know that:

t_down ≈ t_max ≈ 0.738 s

Substituting this value into the equation above, we find:

3.1 ≈ 2 × 0.738 s

Finally, we can use the value of y_max obtained earlier to calculate the height of the building:

y₀ = y_max - v₀y t_down + 1/2 g t_down²

y₀ = y_max - v₀y t_max + 1/2 g t_max²

y₀ ≈ 32.34 m

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