a. The time it takes the ice to reach melting point is: t = 4180 J / 2000 J/s = 2.09 s
b. The time it takes to fully melt is: t = 33300 J / 2000 J/s = 16.65 s
c. The time it takes to reach boiling point after melting is: t = 418000 J / 2000 J/s = 209 s.
Melting Point and Boiling Point Timea. To find the time it takes for the ice to reach its melting point (0°C), we need to calculate the amount of heat required to raise the temperature of the ice block from -10°C to 0°C. The heat required can be calculated using the equation:
Q = m * C * ΔT
where m is the mass of the ice block (100 g)
C is the specific heat capacity of water (4.18 J/g°C)
and ΔT is the change in temperature (0°C - (-10°C) = 10°C).
So, Q = 100 g * 4.18 J/g°C * 10°C = 4180 J
Now, we can use the heat supplied by the stove to calculate the time it takes to raise the temperature of the ice block to 0°C:
t = Q / P
where P is the power supplied by the stove (2000 J/s).
So, t = 4180 J / 2000 J/s = 2.09 s
b. To find the time it takes for the ice to completely melt, we need to calculate the amount of heat required to turn the ice block into liquid water. The heat required can be calculated using the equation:
Q = m * Lf
where m is the mass of the ice block (100 g) and Lf is the latent heat of fusion of water (333 J/g).
So, Q = 100 g * 333 J/g = 33300 J
Now, we can use the heat supplied by the stove to calculate the time it takes to melt the ice block:
t = Q / P
where P is the power supplied by the stove (2000 J/s).
So, t = 33300 J / 2000 J/s = 16.65 s
c. To find the time it takes for the water to reach the boiling point (100°C), we need to calculate the amount of heat required to raise the temperature of the water from 0°C to 100°C. The heat required can be calculated using the equation:
Q = m * C * ΔT
where m is the mass of the water (100 g), C is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature (100°C - 0°C = 100°C).
So, Q = 100 g * 4.18 J/g°C * 100°C = 418000 J
Now, we can use the heat supplied by the stove to calculate the time it takes to raise the temperature of the water to 100°C:
t = Q / P
where P is the power supplied by the stove (2000 J/s).
So, t = 418000 J / 2000 J/s = 209 s.
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A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of the cliff 4.0 s later.
Answer: I am assuming you're asking for the cliff height, which is 24M
We can use the kinematic equations to find the horizontal and vertical components of the ball's velocity and its final position.
The initial velocity components are:
v_x = 33 m/s * cos(60∘) = 28.6 m/s
v_y = 33 m/s * sin(60∘) = 33 * √3/2 m/s
The displacement components can be found using the kinematic equation:
x = v_x * t = 28.6 m/s * 4.0 s = 114.4 m
y = v_y * t - 0.5 * g * t^2 = 33 * √3/2 m/s * 4.0 s - 0.5 * 9.8 m/s^2 * (4.0 s)^2 = 102.4 m - 78.4 m = 24.0 m
The ball lands on the edge of the cliff, so its height must be equal to h, or:
y = h
24.0 m = h
So, the height of the cliff is 24.0 m.
ASAP I WILL GIVE BRAINLEST!!
A 3.0 kg object swings back and forth as a simple pendulum with a small amplitude. The potential energy U of the object as a function of distance x from its equilibrium position is shown above. This particular object has a total energy E of 0.4 J.
b) What is the farthest the object moves along the x-axis in the positive direction? Explain your reasoning.
c) Determine the object’s kinetic energy when its displacement is -8 cm.
d) What is the object’s speed at x = 0?
10cm is the farthest the object moves along the x-axis in the positive direction .0.15J is the object’s kinetic energy when its displacement is -8 cm. 0.316m/s is the object’s speed at x = 0.
Define kinetic energy.
Kinetic energy, which may be seen in the movement of an item or subatomic particle, is the energy of motion. Kinetic energy is present in every particle and moving object. Examples of kinetic energy in action include a person walking, a baseball soaring through the air, a piece of food falling from a table, and a charged particle in an electric field.
Given,
Total energy is 0.4J
m is 3kg
The farthest the object moves along the x-axis in the positive direction would be as potential energy is 0.4J. So from given diagram, displacement will be 10cm
If displacement is -8cm , P.E from diagram will be 0.25J
According to energy conservation formula ,
ME ⇒ U+KE
KE ⇒ ME-U ⇒ 0.4-0.25 ⇒ 0.15J
At x ⇒ 0,
KE ⇒ 1/2 mv^2
0.15 ⇒ 0.5*3*v^2
v ⇒ 0.316m/s
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Below are free-body diagrams for three electric charges that lie in the same plane. Their relative positions are unknown.
Below are free-body diagrams for three electric ch
Along which of the lines (A to H) in the figure(Figure 1)should charge 2 be placed so that the free-body diagrams of charge 1 and charge 2 are consistent?
Figure 2:
Below are free-body diagrams for three electric ch
Along which of the lines (A to H) in the figure(Figure 2)should charge 2 be placed so that the free-body diagrams of charge 1 and charge 2 are consistent?
In Figure 1, charge 2 should be placed along line D to make the free-body diagrams of charge 1 and charge 2 consistent.
The electric field lines produced by charge 2 should point towards charge 1 to provide the attractive force between them. As shown in the free-body diagram of charge 1, the electric field lines point towards the left, which means that charge 2 should be placed on the left side of charge 1. Similarly, in Figure 2, charge 2 should be placed along line F to make the free-body diagrams of charge 1 and charge 2 consistent. The electric field lines produced by charge 2 should point towards the left to provide the attractive force between them. In Figure 1, charge 2 should be placed along line D to make the free-body diagrams of charge 1 and charge 2 consistent.
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Simulate a blackbody spectrum of temperature 1700 Kelvin. Determine the peak wavelength in 3.22 nanometers of an object of that temperature nanometers What is the emissive intensity of the object
a) The peak wavelength in 3.22 nanometers of an object is 345 nanometre, b) the emissive intensity of the object is 2.82 * 10⁸ W/m².
The relationship between the temperature,T and the peak wavelength, [tex]\lambda[/tex] emitted by a black body is given by wien's displacement law:
[tex]\lambda[/tex] = b / T
Where, b is a constant and it's value is 2.898 * 10-3 m-K
Given: T = 8400 K
So, [tex]\lambda[/tex] = (2.898 * 10-3 )/8400
\lambda = 3.45 * 10-7
\lambda = 345 nm
Hence, the peak wavelength of the object at this temperature is 345 nanometre.
The amount of power emitted per unit area, P is given by Stefan Boltzmann law:
P =[tex]\sigma[/tex]T⁴
Where,
Absolute temperature, T = 8400 K
Stefan Boltzmann constant, [tex]\sigma[/tex] = 5.67 * 10-8 W/m²K⁴
So, P = 5.67 * 10-8 * (8400)⁴
P = 2.82 * 10⁸ W/m²
Hence, the power emitted per unit area is 2.82 * 10⁸ W/m².
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12. A truck travelling at 110 km/h has to cover a distance of 640 km in the next 6 hours. Will the driver be on time if he can hold his current speed? (4 marks)
Answer:
calculate the expected time of arrival, we need to divide the total distance by the speed of the truck.
In this case, the expected time of arrival would be:
640 km ÷ 110 km/h = 5.82 hours
So the driver should be able to arrive at the destination within 6 hours if he can hold his current speed.
Suppose the Earth was only half the size it is now (half the mass and half the radius), what would "g" be?
If the Earth were only half the size it is now, the acceleration due to gravity (represented by "g") at the surface would also be halved.
This is because the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them.
With the Earth's mass reduced by a factor of 2 and its radius reduced by a factor of 2, the distance between an object on the surface and the Earth's center would also be reduced by a factor of 2. Thus, the net effect is that the acceleration due to gravity would be halved, resulting in a smaller value of "g" than what we currently observe on Earth.
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A ball is thrown downward with a large starting velocity. Will this accelerate more rapidly than one that is just dropped at the same time?
No, a ball thrown downward with a large starting velocity will not accelerate more rapidly than one that is just dropped at the same time, assuming that both are experiencing the same gravitational field.
Both objects will experience the same b, which is approximately 9.8 m/s^2 near the surface of the Earth. The initial velocity of the thrown ball will only affect its initial speed, but it will not change the acceleration due to gravity.
Therefore, both objects will accelerate at the same rate and will fall at the same speed. However, the thrown ball will cover a greater distance than the dropped ball before hitting the ground, as it has an initial velocity in addition to the acceleration due to gravity.
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A rotating platform with a radius of 2. 0 m makes one complete turn every 3. 0 s. The angular velocity of the platform is most nearly.
In what units do we measure force
Answer:
Newtons
Explanation:
Which of the following are properties of conductors?
I. Holes in the lattice allow the electricity to flow through.
II. Electricity flows easily through this type of material.
III. A few electrons in every atom are loosely held by the nuclei.
A. I only
B. II only
C. I and II
D. II and III
[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]
Lets examine all three properties stated here ~
I) holes in lattice allow the electricity to flow through ?
- holes aren't a majority charge carrier in a conductor, in conductors electricity is conducted by free elecrons. so this statement is incorrect.
ll) Electricity flows easily through this type of material?
- That's true, conductors (usually metals) have free electrons to conduct electricity, which is responsible for good electricity Conductivity.
lll) A few electrons in every atom are loosely held by the nuclei.
- That's also true, Conductors (mainly metals) have a few electrons (say, 1, 2 or maybe 3) in there valence shell which experience quite less force of attraction from nucleus, hence they are free to move around the whole conductor randomly, making a sea of electrons.
So, the correct choice will be : D) ll and lll
Two balls are dropped to the ground from different heights. One is dropped 1.5 s after the other, but they
both strike the ground at the same time, 5.0 s after the first was dropped. (a) What is the difference in the
heights from which they were dropped? (b) From what height was the first ball dropped? Ans. (a) 63 m;
(b) 0.12 km
Answer:
Height difference: approximately [tex]63\; {\rm m}[/tex].
The first ball was dropped from a height of approximately [tex]123\; {\rm m}[/tex].
(Assumptions: both balls were released from rest, air friction is negligible, and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
Under the assumptions, both ball would accelerate at a constant [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex].
Let [tex]t[/tex] denote the time since the first ball was released.
Height of the first ball at time [tex]t[/tex] can be modelled with the SUVAT equation [tex]h(t) = (1/2)\, a\, t^{2} + u\, t + h_{0}[/tex], where [tex]u[/tex] is the initial velocity. However, since [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] by assumption, this equation simplifies to [tex]h(t) = (1/2)\, a\, t^{2} + h_{0}[/tex].
Since this ball reached the ground after [tex]t = 5.0\; {\rm s}[/tex], [tex]h(5.0) = 0\; {\rm m}[/tex]. In other words:
[tex]\begin{aligned}\frac{1}{2}\, (-9.81)\, (5.0)^{2} + h_{0} = 0\end{aligned}[/tex].
Simplify and solve for the initial height of this ball, [tex]h_{0}[/tex]:
[tex]\begin{aligned}h_{0} &= -\frac{1}{2}\, (-9.81)\, (5.0)^{2} \\ &\approx 123\; {\rm m}\end{aligned}[/tex].
In other words, the first ball was dropped from a height of approximately [tex]123\; {\rm m}[/tex].
Similarly, the height of the second ball may be modelled as [tex]h(t) = (1/2)\, a\, t^{2} + h_{0}[/tex].
Since this ball reached the ground [tex]t = (5.0 - 1.5)\; {\rm s} = 3.5\; {\rm s}[/tex] after being released, [tex]h(3.5) = 0\; {\rm m}[/tex]. The initial height of this ball would be:
[tex]\begin{aligned}h_{0} &= -\frac{1}{2}\, (-9.81)\, (3.5)^{2} \\ &\approx (-60)\; {\rm m}\end{aligned}[/tex].
Subtract the initial height of the second ball from that of the first ball to find the difference in initial height:
[tex](123 - 60) \; {\rm m} \approx 63\; {\rm m}[/tex].
A is thrown up into the air with initial vem/s. How fast is the ball moving when it first reachemeters?
Once the ball is thrown, the only force acting on it is gravity, which means that it's acceleration is -9.81 m/s² (negative means downward).
List the known and unknown quantities from the question.
u = initial velocity = 20 m/s
v = final velocity = ? m/s
g = acceleration due to gravity = -9.81 m/s²
t = time interval = ? s
s displacement = 11 m
Before calculating the time it takes for the ball to reach 11 m, the final velocity needs to be calculated using the following kinematic equation.
v² = u² + 2gs
v = √(u² + 2gs)
= √((20 m/s)² + (2x-9.81 m/s² x 11 m)) = 13.57 m/s V=
Calculate the time it takes the ball to reach 11 m using the following kinematic equation.
V = u + gt
Solve for t.
t = (v-u)/g
t (13 57 m/s - 20 m/s)/(-981 m/s²) = 0.655 s
23. A hydraulic (water power) press consists of 1 cm and 5 cm diameter pistons. (a) What force must be applied on the small piston so that the large piston will be able to raise 10 N load? (b) To what height would the load be raised when the small piston has moved 0.1 m?
a. The force required on the small piston to raise a 10 N load on the large piston is 0.403 N.
b. The load would be raised to a height of 63.13 cm when the small piston has moved 0.1 m.
How to calculate?The formula is F1 / F2 = A2 / A1
where F1 is the force applied on the small piston, F2 is the force exerted on the large piston, A1 is the area of the small piston, and A2 is the area of the large piston.
A1 = (π / 4) * (1 cm)^2 = 0.0079 cm^2
A2 = (π / 4) * (5 cm)^2 = 0.196 cm^2
F1 / F2 = 0.0079 cm^2 / 0.196 cm^2 = 0.0403
F2 = 10 N
F1 = F2 * (A1 / A2) = 10 N * 0.0403 = 0.403 N
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The specific heat of copper is 387 J/kg C. The temperature of a 0.35-kg sample of copper decreases from 74.0 °C to 21.0 °C. How much heat flows out of
the copper sample during this temperature drop?
The amount of heat that flows out of the copper sample during this temperature drop is approximately 4,953.75 J.
What is the amount of heat flowing out?
The amount of heat that flows out of the copper sample can be calculated using the formula:
Q = mcΔT
where;
Q is the amount of heat transferred, m is the mass of the copper sample, c is the specific heat of copper, and ΔT is the change in temperature of the sample.Plugging in the given values, we get:
Q = (0.35 kg) x (387 J/kg C) x (74.0 °C - 21.0 °C)
Q = 4,953.75 J
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14. A boy of mass m swings back and forth on the end of a rope 6 m long that is attached to a gymnasium ceiling. If he approaches to within 2 m of the ceiling during each cycle, what is his speed as he passes through the lowest point of the swing?
The boy's speed as he passes through the lowest point of the swing is 8.8 m/s.
How to find speed?To solve this problem, we can use the conservation of energy. At the lowest point of the swing, the boy has the maximum gravitational potential energy and no kinetic energy. At the highest point of the swing, he has the maximum kinetic energy and no gravitational potential energy.
Since the energy of the system is conserved, the sum of the gravitational potential energy and kinetic energy must remain constant throughout the swing. Let's call the speed of the boy as he passes through the lowest point of the swing "v".
At the lowest point, the gravitational potential energy is given by:
mgh = (1/2)mv²
where m is the mass of the boy, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the lowest point above the ground (6 m - 2 m = 4 m).
Solving for v:
v = √(2gh) = √(2 x 9.8 x 4) = √(77.76) = 8.8 m/s
So the boy's speed as he passes through the lowest point of the swing is approximately 8.8 m/s.
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In which collision(s) is momentum conserved?
A.Elastic
B.Inelastic
C.Perfectly inelastic
Momentum is conserved in both elastic and perfectly inelastic collisions.
In an elastic collision, the total momentum of the colliding objects is conserved before and after the collision. This means that the sum of the momentum of the objects before the collision is equal to the sum of the momentum of the objects after the collision.
In a perfectly inelastic collision, the two objects stick together after the collision, forming a single object with new momentum. In this case, the total momentum of the system is also conserved.
However, in an inelastic collision, momentum is not conserved, as some of the momenta are transformed into other forms of energy, such as heat or sound. This means that the total momentum of the objects before the collision is not equal to the total momentum of the objects after the collision.
Answer:
Explanation:
A
A small glass bead has been charged 3 60nCA small metal ball bearing 2.60 cm above the bead feels a 1.8 * 10 2 * N downward electric force
Answer:
The electric field strength at the location of the metal ball bearing is 1.8 * 10^2 N/C.
Newton’s second law of motion states the relationship of mass, acceleration, and force. It says that ______
A.force equals mass divided by acceleration
B. force equals mass multiplied by acceleration
C. acceleration equals force multiplied by mass
D. every object attracts every other object in the universe
Answer:
B. force equals mass multiplied by acceleration
Explanation:
Newton's second law of motion states that the acceleration of an object equals the net force acting on the object divided by the object's mass. According to the second law, there is a direct relationship between force and acceleration and an inverse relationship between mass and acceleration.
Newton's second law of motion is F = ma, or force is equal to mass times acceleration.
TRUE OR FALSE: Increasing the time of impact during a collision increases the force experienced by the object. * True False
Answer:
False
Explanation:
I'm pretty sure increasing the time of impact actually decreases the force because it is being spread out.
You will now use the observations you have made so far to predict what the strength of the field will be at point (0, 50).A. Using the right-hand rule, which direction is the magnetic field at (0, 50)?B. Since the magnetic field of the Earth and the induced field are at right angles, you can use the Pythagorean Theorem to determine the strength of the combined field.
In A. part, the magnetic field at (0, 50) is in west direction. In B. part, the strength of the field at (0,50) is 2.06 G.
A. The current is flowing up for west as shown by the front view figure at the bottom of the gadget. Your fingers will curve to the west if you wrap your right hand around the wire with your thumb up. Put a compass at (0,50) to check the direction as well. It indicates west.
B. By using the Pythagorean Theorem to determine the strength of the combined field, the strength of the field at (0,50) is 2.06 G.
The earth's magnetic field strength= 0.50 G
The induced current magnetic field strength= 2.0
B is given by=
[tex]\sqrt{0.50^{2} - 2.00^{2} }\\ =\sqrt{0.25-4.00}\\ =2.06[/tex]
Hence, we can also check by putting the probe on (0,50) and the probe reads 2.06 G.
Therefore, the strength of the field at (0,50) is 2.06 G.
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You are asked to relocate a sunken boat to be used as an artificial reef. The boat's mass is 4,000Kg and the volume of its solid parts is 3m3. Assume the ocean water density is 1.02g/ml. Approximately, what is the minimal volume of an air bag (in liters) that should be attached to it so it can be lifted under water to be moved to its new location (the air bag weighs 10Kg)?
Without having the help of the moving pad, you and a friend have to move the very heavy
entertainment center. The kinetic friction opposing the motion is equal to 1600 N and your
maximum applied force is 1000N. How much force does your friend need to apply to make the
entertainment center move at a constant velocity?
The force needed by my friend to make the entertainment center move at a constant velocity is 600 N
How to find force application?To find the force application in a specific situation, one needs to consider the forces that are acting on an object and the net force (the vector sum of all forces acting on an object) which determines the motion of the object.
In order to make the entertainment center move at a constant velocity, the sum of the applied forces must equal the force of kinetic friction. If the maximum applied force you can provide is 1000 N, then your friend needs to apply a force of 1600 N - 1000 N = 600 N.
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5. On some highways, exit signs are numbered according to the number of kilometres the
exit is from the place where the highway originated. If you are driving south and pass
exit 35 at 2:15 pm and then you pass exit 116 at 3:09 pm, what is your velocity in m
Answer:
We can start by using the formula:velocity = distance/timeFirst, we need to calculate the distance traveled in kilometers.
To do this, we can subtract the exit numbers:
116 km - 35 km = 81 km
Next, we need to convert the time difference from hours and minutes to hours:
3:09 pm - 2:15 pm = 0.9 hours
Now we can use the formula to find the velocity:
velocity = 81 km / 0.9 hours
velocity ≈ 90 km/h
Finally, we can convert this velocity to meters per second by multiplying by 1000/3600:
velocity = 90 km/h x 1000 m/km / 3600 s/h
velocity ≈ 25 m/s
Therefore, your velocity is approximately 25 m/s.
Explanation:
4 Carbon monoxide gas (CO) contained within a piston–
cylinder assembly undergoes three processes in series:
Process 1–2: Constant pressure expansion at 5 bar from V1 5
0.2 m3
to V2 5 1 m3
.
Process 2–3: Constant volume cooling from state 2 to state 3
where p3 5 1 bar.
Process 3–1: Compression from state 3 to the initial state during
which the pressure–volume relationship is pV = constant.
Sketch the processes in series on p–V coordinates and
evaluate the work for each process, in kJ.
To sketch the processes in a p-V diagram, we need to first determine the initial and final states of each process, as well as the path each process takes.
How do we determine the state of each process?Process 1-2 is a constant pressure expansion from state 1 to state 2. So, the path is a straight horizontal line on the p-V diagram, from (0.2, 5) to (1, 5) (in units of m^3 and bar).
Process 2-3 is a constant volume cooling from state 2 to state 3, so the path is a straight vertical line on the p-V diagram, from (1, 5) to (1, 1).
Process 3-1 is a compression process during which the pressure-volume relationship is pV=constant. This implies that the path on the p-V diagram is a hyperbola, passing through state 3 and returning to state 1.
The work done in each process can be calculated using the following equations:
W = P(V2 - V1) for constant pressure process (1-2)
W = 0 for constant volume process (2-3)
W = -nRT ln(V2/V1) for isothermal process (3-1), where n is the number of moles of CO, R is the gas constant, and T is the temperature of the gas.
Assuming standard temperature and pressure conditions (STP), which is 1 atm and 273.15 K, the gas constant R can be taken as 0.0821 Latm/(molK).
Using these equations, we can calculate the work for each process as follows:
W1-2 = 5*(1-0.2) = 4 kJ
W2-3 = 0
W3-1 = -nRT ln(V2/V1) = -10.0821273.15 ln(1/0.2) = 11.1 kJ
Therefore, the total work done on the gas in the three processes is the sum of the work done in each process, which is 4 kJ + 0 kJ + 11.1 kJ = 15.1 kJ.
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Let E-> =3i^+1j^ and F -> =1i^−3j^. (Note: the -> means it is above the letter!)
A. Find the magnitude of E -> .
B. Find the magnitude of F ->.
C. Find the magnitude of G -> = E-> +F -> .
D. Find the magnitude of H-> =−E -> −2F -> .
A. the magnitude of E -> is √10.
B. the magnitude of F -> is √10
C. the magnitude of G -> = E-> +F -> is √20
D. the magnitude of H-> =−E -> −2F -> is √50.
What is magnitude?The magnitude or size of a mathematical object is described as a property which determines whether the object is larger or smaller than other objects of the same kind.
A. The magnitude of the vector E -> is given by the formula:
|E -> | = √(3^2 + 1^2) = √(9 + 1) = √10
So, the magnitude of E -> is √10.
B. The magnitude of the vector F -> is given by the formula:
|F -> | = √(1^2 + (-3)^2) = √(1 + 9) = √10
So, the magnitude of F -> is √10.
C. The magnitude of the vector G -> is given by the formula:
G -> = E-> + F ->
|G -> | = √((3 + 1)^2 + (1 - 3)^2) = √(4^2 + (-2)^2) = √(16 + 4) = √20
So, the magnitude of G -> is √20.
D. The magnitude of the vector H -> is given by the formula:
H -> = -E-> - 2F->
|H -> | = √((-3 - 2 * 1)^2 + (-1 - 2 * -3)^2) = √((-5)^2 + 5^2) = √50
So, the magnitude of H -> is √50.
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In the experiment that student needs to know the mass of four different objects: a shoe, piece of paper,book, and a tissue.which of the following units of measurement should the student use
The unit of measurement the student should use is the gram. The correct option is c.
What are the measurements units?
The International System of Units (SI) uses the gram (formerly gram; SI unit symbol g) as the unit of mass that corresponds to one-thousandth of a kilogram.
A student must know the masses of four different objects for an experiment: a shoe, a piece of paper, a book, and a tissue. The learner should utilize grams as their measurement units.
Therefore, the correct option is c. gram is used to measure the mass of objects and things.
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The question is incomplete. The missing options are given below:
milligram
ounce
gram
kilogram
A 0.0272 kg arrow is fired with a velocity of 87.3 m/s from a 0.0245 kg bow. Determine the equivalent impulse value imparted on the arrow. (hint: you can also use newton's third law of motion)* A.2.37 Ns B.3.11 Ns C.113.5 Ns D.214.2 Ns
Answer:
Explanation:
C. 113.5 Ns
Unit 3
The Flow of Energy in Systems
EVIDENCE NOTEBOOK
11. How could the surfaces and structures in an urban area be related to the
urban heat island effect? Record your evidence
Answer:
The surfaces and structures in an urban area can have a significant influence on the urban heat island effect. For example, paved surfaces like roads and buildings absorb more heat than surfaces covered in vegetation, and structures like high-rise buildings can trap and reflect heat, resulting in higher temperatures in the urban area. Additionally, urban areas usually have less vegetation than their rural counterparts, meaning there is less vegetation to absorb heat from the sun and provide shade, further contributing to the urban heat island effect. Evidence of this can be seen in research at louisvilleky.gov/government/sustainability/urban-heat-island-project.
Calculate the total work done on a mass m as it moves from position x1 = 0 m to x2 = 40 m
F1 = 5 N F2 = 6 N angle between them 50 degree
F3 = 2 N F4 = 2 N Angle Between them 20 Degree
The total work done on the mass as it moves from x1 = 0 m to x2 = 40 m is approximately 515.17 J.
What is Work Done?
Work is a physical quantity that describes the amount of energy transferred when a force acts on an object and causes it to move. When a force acts on an object and causes it to move in the direction of the force, work is said to be done on the object. Mathematically, work is defined as the dot product of force and displacement:
Work = Force x Displacement x cos(theta)
To calculate the total work done on the mass as it moves from position x1 to x2, we need to find the net work done by all the forces on the mass. The net work done by a force is given by the formula:
W = F * d * cos(theta)
where W is the work done, F is the force, d is the displacement of the mass, and theta is the angle between the force and the displacement.
First, we can calculate the work done by each force separately and then add them up to find the total work done.
Work done by F1:
W1 = F1 * (x2 - x1) * cos(0) = 5 N * 40 m * cos(0) = 200 J
Work done by F2:
W2 = F2 * (x2 - x1) * cos(50°) = 6 N * 40 m * cos(50°) ≈ 165.41 J
Work done by F3:
W3 = F3 * (x2 - x1) * cos(20°) = 2 N * 40 m * cos(20°) ≈ 74.88 J
Work done by F4:
W4 = F4 * (x2 - x1) * cos(20°) = 2 N * 40 m * cos(20°) ≈ 74.88 J
The total work done on the mass is the sum of the work done by each force:
W_total = W1 + W2 + W3 + W4 ≈ 515.17 J
Therefore, the total work done on the mass as it moves from x1 = 0 m to x2 = 40 m is approximately 515.17 J.
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A motorcycle moves according to the velocity-versus-time graph shown in the figure
Find the average acceleration of the motorcycle during segment A of the motion.
(part a)
Find the average acceleration of the motorcycle during segment B of the motion.
(part b)
Find the average acceleration of the motorcycle during segment C of the motion.
(part c)
Average acceleration A = 2 m/s² Average acceleration B = 0 m/s² Average acceleration C = -0.5 m/s²
What is acceleration?The rate of change in velocity over time is called acceleration. The unit of measurement for this vector quantity is meters per second squared (m/s²). Acceleration can be either positive (speeding up) or negative (speeding down). It can also be referred to in terms of direction, such as acceleration to the left or right.
a) V=10m/s
u=0m/s
t= 5 second
a = (v-u)/t
= (10-0)/5
= 2 m/s²
b) The body moves with constant velocity 10m/s , so acceleration is 0m/s
c The velocity is falling , so the body is,
a= (v-u)/t= (5-10)/10
= -5/10
= -0.5 m/s²
Therefore, the Average acceleration of A, B and C are 2 m/s², 0m/s and -0.5 m/s²
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