what type of force

a child on a sled slides down the hill

The angle  made with the ceiling by the tension force of the two wires is determined as 50⁰.

The angle  made by the two tensions is calculated by applying cosine rule as follows;

the force opposite the angle = weight of the block = mg

W = 7.75 kg x 9.8 m/s²

W = 75.95 N

The angle  made by the two tensions is calculated as follows;

cos θ = (38 N ) / ( 59 N)

cos θ = 0.6441

θ = arc cos (0.6441)

θ = 50⁰

Thus, the angle  made by the two tensions is determined as 50 degrees.

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The figure is in the image attached

The brightness of light is explained by the wave model of light. Brightness refers to the perceived intensity of light. Brightness is determined by the amplitude or intensity of light waves.

The larger the amplitude, the brighter the light. This can be explained using the wave model of light.Light is a form of electromagnetic radiation that is composed of oscillating electric and magnetic fields. The wave model of light states that light is a transverse wave that propagates through space. The wave model of light states that light travels in straight lines and can be reflected, refracted, and diffracted. Brightness is a measure of the intensity of light waves.

The intensity of light waves is determined by the amplitude of the wave.The amplitude of a wave is the maximum displacement of the wave from its equilibrium position. The larger the amplitude of a wave, the more energy the wave carries. This means that the larger the amplitude of light waves, the brighter the light. The brightness of light can be increased by increasing the amplitude of light waves. This can be achieved by increasing the intensity of light waves. The intensity of light waves can be increased by increasing the power of the light source. Thus, brightness can be explained by the wave model of light as it is determined by the amplitude or intensity of light waves.

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If two waves with equal amplitudes and wavelengths travel through a medium in such a way that a particular particle of the medium is at the crest of one wave and at the trough of the other wave at the same time then D) The particle will remain stationary due to interference.

When two waves with equal amplitudes and wavelengths pass through a medium, they undergo interference. Interference occurs when the crests and troughs of the waves overlap. In this case, if a particular particle of the medium is at the crest of one wave and at the trough of the other wave at the same time, it experiences what is called destructive interference.

Destructive interference happens when the peaks (crests) of one wave align with the troughs of the other wave. In this situation, the positive displacements caused by the crest are canceled out by the negative displacements caused by the trough. As a result, the net displacement of the particle is zero, and it remains stationary.

This phenomenon occurs due to the principle of superposition, which states that the total displacement of a particle at any point in a medium is the vector sum of the individual displacements caused by each wave. Therefore, in this scenario, the particle will remain stationary due to the destructive interference between the two waves. Therefore, Option D is correct.

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Answer: D

Explanation:

a. The maximum height reached by the rocket is  1083 meters.

b.  The rocket reaches its maximum height 38 seconds after liftoff.

c.   The rocket is in the air for  1.09 seconds.

(b)

We will apply equation of motion :

v² = u² + 2aΔy

Δy = (v² - u²) / (2a)

Δy = (0 - 57.0²) / (2 * 1.50)

Δy = (-57.0)² / 3.00

Δy = 3,249 / 3.00

Δy = 1083 m

(c)

v = u + at

0 = u + at

t = -u / a

t = -57.0 / 1.50

t = 38 seconds

(d)

Δy = ut + (1/2)at²

140 = 57.0t + (1/2)(1.50)t²

(1/2)(1.50)t² + 57.0t - 140 = 0

t =  1.09 seconds.

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(a) the magnitude of the displacement is approximately 3.85 km, and the direction is approximately 59.04° south of east.

(b) Magnitude of average velocity ≈ 3.85 km and Direction of average velocity ≈ -59.04° (south of east)

(c) the average speed during the given time interval is approximately 1.66 km/h.

(a) To find the magnitude and direction of the person's displacement, we can use the Pythagorean theorem and trigonometry.

Displacement in the x-direction = 2.00 km east

Displacement in the y-direction = -3.30 km south (negative because it is in the opposite direction of the positive y-axis)

Using the Pythagorean theorem:

Magnitude of displacement = √((2.00 km)^2 + (-3.30 km)^2)

Magnitude of displacement ≈ 3.85 km

To find the direction, we can use trigonometry:

θ = tan^(-1)(opposite/adjacent)

θ = tan^(-1)(-3.30 km / 2.00 km)

θ ≈ -59.04° (measured counterclockwise from the positive x-axis)

Therefore, the magnitude of the displacement is approximately 3.85 km, and the direction is approximately 59.04° south of east.

(b) Average velocity is defined as displacement divided by time. The magnitude and direction of average velocity will be the same as the magnitude and direction of displacement.

Magnitude of average velocity ≈ 3.85 km

Direction of average velocity ≈ -59.04° (south of east)

(c) Average speed is defined as total distance traveled divided by time. The total distance traveled is the sum of the magnitudes of the individual displacements.

Total distance = 3.30 km + 2.00 km = 5.30 km

Average speed = Total distance / Time

Average speed ≈ 5.30 km / 3.20 hours

Average speed ≈ 1.66 km/h

Therefore, the average speed during the given time interval is approximately 1.66 km/h.

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(a) The amplitude of the wave is 0.2 m.

(b) The period of the wave is  4 s.

(c) The wavelength of the wave is 100 m.

(a) The amplitude of the wave is the maximum displacement of the wave.

amplitude of the wave = 0.2 m

(b) The period of the wave is the time taken for the wave to make one complete cycle.

period of the wave = 5.5 s - 1.5 s = 4 s

(c) The wavelength of the wave is calculated as follows;

λ = v / f

where;

f = 1/t = 1 / 4s = 0.25 Hz

λ = ( 25 m/s ) / 0.25 Hz

λ = 100 m

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(a) The average velocity of the tennis player at 0 to 1 s is  4 m/s.

(b) The average velocity of the tennis player at 0 to 4 s is -0.5 m/s.

(c) The average velocity of the tennis player at 1 to 5 s is  1 m/s.

(d) The average velocity of the tennis player at 0 to 5 s is 0.8 m/s.

The average velocity of the tennis player at the given time, is calculated by applying the formula for average velocity as follows;

average velocity = total displacement / total time

(a) The average velocity at 0 to 1 s;

average vel. = (4 m - 0 m ) / (1 s ) = 4 m/s

(b) The average velocity at 0 to 4 s;

average vel. = (-2 - 0 )m / 4 s = -0.5 m/s

(c) The average velocity at 1 to 5 s;

average vel. = (4 - 0 )m / (5 - 1) s =  1 m/s

(d) The average velocity at 0 to 5 s;

average vel. = (4 - 0 )m / (5 - 0) s =  0.8 m/s

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The force acting on the electron is 1.92 x 10^-17 N.

The problem states that an electron of charge 1.6 x 10^-19 is located in a uniform electric field of 120 Vm^-1, and it asks us to determine the force acting on it.

We can use Coulomb's law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. If the charges are of opposite signs, the force is attractive, while if the charges are of the same sign, the force is repulsive.

The formula for Coulomb's law is F = kq1q2/r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.

Since the electron has a charge of 1.6 x 10^-19 C, and the electric field strength is 120 Vm^-1, we can use the equation F = qE to find the force acting on it.

F = qE = (1.6 x 10^-19 C)(120 Vm^-1) = 1.92 x 10^-17 N.

Therefore, the force acting on the electron is 1.92 x 10^-17 N.

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(a) The projectile's velocity at the highest point of its trajectory is approximately 27.01 m/s, counterclockwise from the +x-axis.

(b) The straight-line distance from where the projectile was launched to where it hits its target is approximately 144.93 m.

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

(a) To find the projectile's velocity at the highest point of its trajectory, we need to consider the vertical component and horizontal component separately.

The initial velocity (V0) of the projectile is 48.0 m/s, and the launch angle (θ) is 34.0° above the horizontal.

The vertical component of velocity (Vy) can be found using the equation:

Vy = V0 * sin(θ)

Plugging in the known values:

Vy = 48.0 m/s * sin(34.0°)

Calculating Vy, we find:

Vy ≈ 27.01 m/s

The horizontal component of velocity (Vx) can be found using the equation:

Vx = V0 * cos(θ)

Plugging in the known values:

Vx = 48.0 m/s * cos(34.0°)

Calculating Vx, we find:

Vx ≈ 39.79 m/s

Therefore, at the highest point of its trajectory:

- The magnitude of the projectile's velocity is approximately 27.01 m/s.

- The direction of the velocity is straight up, counterclockwise from the +x-axis.

(b) To find the straight-line distance from where the projectile was launched to where it hits its target, we need to consider the horizontal motion of the projectile.

The time of flight (t) is given as 3.65 s.

The horizontal distance (x) can be found using the equation:

x = Vx * t

Plugging in the known values:

x = 39.79 m/s * 3.65 s

Calculating x, we find:

x ≈ 144.93 m

The straight-line distance from where the projectile was launched to where it hits its target is approximately 144.93 m.

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Answer:

3.79 m/s^2

Explanation:

We know the small part loses 0.932 MJ of gravitational potential energy during its fall.

Potential energy = mass x gravitational field strength x height

Re-arranging to solve for gravitational field strength:

g = Potential energy/(mass x height)

Plugging in the given values:

g = 0.932 MJ / (1.8kg x 140km)

= 0.932 x 10^6 J / (1.8 x 1000kg x 140 x 1000m)

= 3.79 m/s^2

Therefore, the gravitational field strength of Mars is calculated to be 3.79 m/s^2.

The second snowball should be thrown at an angle of approximately 48.196° with respect to the horizontal to arrive at the same point as the first snowball.

the second snowball should be thrown 4.582 seconds later in order for both to arrive at the same time.

To find the angle at which the second snowball should be thrown, we can use the fact that the horizontal displacement of both snowballs must be the same.

Let's first find the horizontal and vertical components of the velocity for the first snowball. The initial speed is 26.5 m/s, and the angle is 58.0° with respect to the horizontal.

The horizontal component of the velocity for the first snowball is given by:

V1x = V1 * cos(angle1)

    = 26.5 m/s * cos(58.0°)

    = 26.5 m/s * 0.530

    = 14.045 m/s

Now, let's find the vertical component of the velocity for the first snowball:

V1y = V1 * sin(angle1)

    = 26.5 m/s * sin(58.0°)

    = 26.5 m/s * 0.848

    = 22.472 m/s

Since the vertical acceleration is the same for both snowballs (gravity), the time it takes for both to arrive at the same point is the same. Therefore, we can use the time of flight of the first snowball to calculate the vertical displacement for the second snowball.

The time of flight for the first snowball can be calculated using the vertical component of velocity and the acceleration due to gravity:

t = (2 * V1y) / g

  = (2 * 22.472 m/s) / 9.8 m/s²

  ≈ 4.582 s

Now, let's find the vertical displacement for the second snowball:

Δy = V1y * t - (0.5 * g * t²)

    = 22.472 m/s * 4.582 s - (0.5 * 9.8 m/s² * (4.582 s)²)

    ≈ 103.049 m

To find the angle at which the second snowball should be thrown, we can use the horizontal displacement and the vertical displacement:

tan(angle2) = Δy / Δx

           = 103.049 m / (2 * 14.045 m/s * t)

           = 103.049 m / (2 * 14.045 m/s * 4.582 s)

           ≈ 1.085

Now, we can find the angle2 by taking the arctan of both sides:

angle2 ≈ arctan(1.085)

angle2 ≈ 48.196°

Therefore,

To find how many seconds later the second snowball should be thrown, we can simply use the time of flight of the first snowball, which is approximately 4.582 seconds.

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