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My name is Ann [436]
3 years ago
12

A sled and rider with a combined mass of 70 kg are at the top of a hill that rises 9 m above the level ground below. The sled is

given a push, providing an initial kinetic energy at the top of the hill of 1,200 J. a. Choosing a reference level at the bottom of the hill, what is the potential energy of the sled and rider at the top of the hill? sled and rider at the top of the hill? of the sled and rider at the bottom of the hill? b. After the push, what is the total mechanical energy of the hill?c. If friction can be ignored, what will be the kinetic energy of the sled and rider at the bottom of the hill?

Physics
1 answer:
kvv77 [185]3 years ago
3 0

Answer:

(A) 6174J, 0J.

(B) 7374J

(C) 7374J

Explanation:

See the attachment below for the calculation.

The princi6of conservation of energy has been used.

E = K1 + U1 = K2 + U2

Where E = total mechanical energy.

Taking the bottom as reference, and h =0 and as a result U2 = mg(0) = 0J.

The complete solution to the problem can be found in the attachment below.

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Describe two experiments to determine the speed of propagation of a transverse wave on a rope. You have the following tools to u
AnnZ [28]

Answer:

#See solution for details.

Explanation:

1.

Tools:stopwatch, \ meter \ stick, \ mass \ measuring \ scale , \ force \ measuring  \ device.

Experiment \ 1:Calculate the speed of the wave using the time,t it takes to travel along the rope. Rope's length,L is measured using the meter stick.

-Attach one end of rope to a wall or post, shake from the unfixed end to generate a pulse. Measure the the time it takes for the pulse to reach the wall once it starts traveling using the stopwatch.

-Speed of the pulse can then be obtained as:

v=\frac{L}{t}

Experiment \ 2: Apply force of known value to the rope then use the following relation equation to find the speed of a pulse that travels on the rope.

v=\sqrt{\frac{F}{\mu}}\ ,\mu=\frac{m}{L}

-Use the measuring stick and measuring scale to determine L,m values of the rope then obtain \mu.

-Use the force measuring constant to determine F. These values can the be substituted in Experiment \ 1 to obtain v.

4 0
3 years ago
The generator at a power plant produces AC at 20,000 V. A transformer steps this up to 355,000 V for transmission over power lin
Masja [62]

Answer:

Number of coil in the output is 39938

Explanation:

We have given a step up transformer

Input voltage of transformer, that is primary voltage v_p=20000volt

Output voltage, that is secondary voltage v_s=355000volt

Number of turns in primary N_p=2250

For transformer we know that \frac{V_p}{V_s}=\frac{N_p}{N_s}

\frac{20000}{355000}=\frac{2250}{N_s}

N_s=39937.5

As the number of turns can not be in fraction so number of turns in the output coil is 39938

7 0
3 years ago
A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher
Scilla [17]

Answer:

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

Explanation:

The Coulomb's Law gives the force by the charges:

\vec{F} = K\frac{q_1q_2}{r^2}\^r

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)

where Θ is the angle between F_1 and x-axis.

F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

whereas

F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

Finally, the x-component of the net force is

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

8 0
3 years ago
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