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Pie
4 years ago
9

A dog of mass 10 kg sits on a skateboard of mass 2 kg that is initially traveling south at 2 m/s. The dog jumps off with a veloc

ity of 1 m/s north rela-tive to the ground. Which of the following is the best estimate of the velocity of the skateboard immediately after the dog has jumped?
(A) 1 m/s north(B) 1 m/s south(C) 3 m/s south (D) 7 m/s south(E) 17 m/s south
Physics
1 answer:
Tasya [4]4 years ago
3 0

Answer:

17 m/s south

Explanation:

m_1 = Mass of dog = 10 kg

m_2 = Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

u_1 = Velocity of dog = 1 m/s

u_2 = Velocity of skateboard

In this system the linear momentum is conserved

(m_1+m_2)v+m_1u_1+m_2u_2=0\\\Rightarrow u_2=-\dfrac{(m_1+m_2)v+m_1u_1}{m_2}\\\Rightarrow u_2=-\dfrac{(10+2)2+10\times 1}{2}\\\Rightarrow u_2=-17\ m/s

The velocity of the skateboard will be 17 m/s south as the north is taken as positive

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3 years ago
The hockey player is moving at a speed of 9. 5 m/s. if it takes him 2 seconds to come to a stop under constant acceleration, how
Lunna [17]

Answer:

9.5\; {\rm m}.

Explanation:

Let u and v denote the velocity of this hockey player before and after stopping, respectively. The question states that u = 9.5\; {\rm m\cdot s^{-1}} and implies that v = 0\; {\rm m\cdot s^{-1} since the hockey player has come to a stop.

The duration of this acceleration is t = 2\; {\rm s}.  

Since the acceleration of this hockey player was constant, SUVAT equation would apply. In particular, the SUVAT equation x = (1/2)\, (v + u) \, (t) gives the displacement x of this hockey player during that 2\; {\rm s} of acceleration:

\begin{aligned} x &= \frac{1}{2}\, (9.5\; {\rm m\cdot s^{-1}} + 0\; {\rm m\cdot s^{-1}})\, (2\; {\rm s}) = 9.5\; {\rm m} \end{aligned}.

In other words, this hockey player would have travelled 9.5\; {\rm m} while stopping.

8 0
2 years ago
Two wires are made of the same material. Wire 1 has length that is 1.35 times the length of wire 2 and diameter that is 0.91 tim
Oksi-84 [34.3K]

Answer:

1.117935:1

Explanation:

Since the wires are of the same material, they will have the same resistivity \rho.

The cross-sectional area of the of a wire is given by;

A=\pi\frac{d^2}{4}................(1)

where d is the diameter of the wire.

Also, the relationship between resistance R, cross-sectional area A and length l of a wire is given as follows;

\rho=\frac{RA}{l}..................(2)

Since the resistivity same for both wires, say wire 1 and wire 2, we can wreite the following;

\frac{R_1A_1}{l_1}=\frac{R_2A_2}{l_2}..................(3)

Hence from eqaution (3), the ration of wire 1 to 2 is expressed as;

\frac{R_1}{R_2}=\frac{l_1A_2}{l_2A_1}..................(4)

Given;

l_1=1.35l_2

\frac{R_1}{R_2}=\frac{1.35l_2A_2}{l_2A_1}\\\frac{R_1}{R_2}=\frac{1.35A_2}{A_1}.............(5)

We then use equation (1) to fine the ratio of the area A_1 to A_2

bearing in mind that d_1=0.91d_2

This ratio gives 0.8281. Substituting this into equation (5), we get the following;

\frac{R_1}{R_2}= 1.35*0.8281=1.117935

4 0
3 years ago
Read 2 more answers
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