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3 years ago

A child carries a 3N book at a constant velocity of 4 meters across a horizontal floor. What is the net work done?

1 answer:

s2008m [1.1K]3 years ago

6 0

This problem is getting you used to the definition of 'work' = (force) x (distance).

It's hard to really imagine, but technically, when an object moves horizontally,
no work is done, because there's no lifting against the force of gravity.

Another way to look at it: The potential energy of the object never changes,
because its height never changes, so no energy ever goes into it ot gets
taken out of it.

Also, if the object moves at a constant speed, then it never accelerated,
and that means that no force ever acted on it.

So technically, when an object moves horizontally at a constant speed,
the work done on it is zero.

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Fantom [35]

Answer:

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Explanation:

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and since temperature in above block is less than the lower block, the heat will flow from lower block to higher block .

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3 years ago

Waves combine to produce a smaller or zero-amplitude wave in a process called

vampirchik [111]

ANSWER: Destructive Interference

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Hope this Helps!

6 0

3 years ago

Read 2 more answers

An object has an angular velocity. It also has an angular acceleration due to torques that are present. Therefore, the angular v

shusha [124]

Answer:

α = 0 , w = w₀

Explanation:

Torque is related to angular acceleration by Newton's second law for rotational motion.

τ = I α

Where τ is the torque, I the moment of inertia and α the angular acceleration.

If we apply an external torque for the sum of all torques to be zero, the angular acceleration must fall to zero

α = 0

Since the acceleration is zero, the angular velocity you have at that time is constantly killed.

w = w₀ + α t

w = w₀ + 0

8 0

3 years ago

If the escalator pulls you up a slope of 30.0o and moves with a velocity of 3.10m/s. What is the vertical component of your v

ioda

Answer:

$1.55\ \text{m/s}$

$2.68\ \text{m/s}$

Explanation:

v = Velocity of the elevator = 3.1 m/s

$\theta$ = Angle of the slope = $30^{\circ}$

Vertical component is given by

$v_y=v\sin\theta\\\Rightarrow v_y=3.1\sin30^{\circ}\\\Rightarrow v_y=1.55\ \text{m/s}$

The vertical component of the velocity is $1.55\ \text{m/s}$ .

Horizontal component is given by

$v_x=v\cos\theta\\\Rightarrow v_x=3.1\times \cos30^{\circ}\\\Rightarrow v_x=2.68\ \text{m/s}$

The horizontal component of the velocity is $2.68\ \text{m/s}$ .

8 0

3 years ago

A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t

Reil [10]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

7 0

3 years ago