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trapecia [35]
2 years ago
5

How much work is done moving an object 5.0 m by a 15 N force? a. 75 J b. 15 J c. 5.0 J d. 3.0 J

Physics
1 answer:
White raven [17]2 years ago
7 0

Answer:

75 Joules

Explanation:

Work=F*D, plugging in the constant force of 15 newtons multiplying it by the displacement of 5 meters, it results in 75 Joules of work.

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What is the magnitude of the net torque on the pulley about the axle?
jeyben [28]

Answer: Torque= tangential force × radius of the pulley

Explanation:

Pulley is a simple machine which helps us to lift loads by applying the force in a convenient direction. It consists of a grooved wheel which is free to rotate about an axle passing through the center of the pulley.

For a pulley, the torque T= tangential force × radius of the pulley

Pulleys can be used in combination also to achieve a desired output.

6 0
3 years ago
Which vector best represents the net force acting on the +3 C charge
adelina 88 [10]
Vector ' W ' best and there ya go
7 0
3 years ago
What are the characteristics of high energy wave?
Ivahew [28]

Answer:

D. High frequency and short wavelengths.

Explanation:

If a wave is high in energy it will have a higher frequency.

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8 0
2 years ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
A shot is fired at an angle of 60 degree horizontal with Kinetic energy E. If air resistance is ignored, the K.E at the top of t
Lapatulllka [165]
I'm not sure what "60 degree horizontal" means.

I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith. 

Now, I'll answer the question that I have invented.

When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is    S cos(60)  =  0.5 S ,
and the vertical component is   S sin(60) = S√3/2  =  0.866 S .  (rounded)

-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.

-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change. 

-- So at the top of its trajectory, its KE is 0.25 of what it had originally. 

That's  E/4 .
3 0
3 years ago
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