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beks73 [17]
3 years ago
5

A child pulls on a wagon with a force of 75 N. If the wagon moves a total of 42 m in 3.1 min, what is the average power delivere

d by the child
Physics
1 answer:
exis [7]3 years ago
4 0

Answer:

16.96 W

Explanation:

Power: This can be defined as the rate at which work is done by an object. The S.I unit of power is Watt(W).

From the question,

P = (F×d)/t....................... Equation 1

Where P = power, F = force, d = distance, t = time.

Given: F = 75 N, d = 42 m, t = 3.1 min = 3.1×60 = 186 s

Substitute these values into equation 1

P = (75×42)/186

P = 16.94 W

Hence the average power delivered by the child  = 16.96 W

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A car traveling initially at 7.35 m/s acceler-
Flura [38]

Answer:

v_f=9,07~m/s

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes uniformly in time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

v_f=v_o+at

The car initially travels at vo=7.35 m/s and accelerates at a rate of a=0.824~m/s^2 during t=2.09 s.

The final velocity is:

v_f=7.35+0.824*2.09

\mathbf{v_f=9,07~m/s}

4 0
2 years ago
the diagram below shows the situation described in the problem. the focal length of the lens is labeled f; the scale on the opti
Artist 52 [7]

Answer:

I have no clue

Explanation:

8 0
2 years ago
What component of a longitudinal sound wave is analogous to a trough of a transverse wave?
anzhelika [568]

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.  

6 0
3 years ago
An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera
Ede4ka [16]
The amount of heat given by the water to the block of ice can be calculated by using
Q=m_w C_{sw} \Delta T_w
where 
m_w = 30 g is the mass of the water
C_{sw}=4.18 J/(g ^{\circ}C) is the specific heat capacity of water
\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C is the variation of temperature of the water.

Using these numbers, we find
Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
Q = m_i L_f
where m_i is the mass of the ice while L_f =334 J/g is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
m_i =  \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g
3 0
3 years ago
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
hammer [34]
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
3 years ago
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