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Cerrena [4.2K]
3 years ago
9

A mother and daughter press their hands together and then push apart while ice skating. Immediately after they push away from ea

ch other, how does the motion of the mother and daughter change? ​

Physics
2 answers:
kherson [118]3 years ago
7 0

Answer:

J

Explanation:

The daughter moves with greater acceleration backwards because of her weight.

vova2212 [387]3 years ago
3 0

Answer:D on DCP

Explanation:

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A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
stealth61 [152]

Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

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3 years ago
Which of the following is true regarding the speed of earthquake waves?
Levart [38]

Answer:

p waves travel faster than s waves and surface waves

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3 years ago
A force of 6.00 N acts in the positive direction on a 3.00 kg object, originally traveling at +15.0 m/s, for 10.0 s. (a) What is
frozen [14]

Answer:

60 kg m/s

Explanation:

Let a\;\; m/s^2 be the acceleration of the object.

As the acceleration of the object is constant, so

a=\frac {v-u}{t}\cdots(i)

Given that applied force, F=6.00 N,

From Newton's second law, we have

F= m\times a,

\Rightarrow F=\frac {m(v-u)}{t} [from equation (i)]

\Rightarrow Ft=m(v-u)

\Rightarrow Ft=mv-mu

\Rightarrow mv-mu=6\times 10 [given that time, t=10 s and F=6 N]

\Rightarrow mv-mu=60 kg \;m/s

Here mv is the final momentum of the object and mu is the initial momentum of the object.

So, the change in the momentum of the object is mv-mu.

Hence, the change in the momentum of the object is 60 kg m/s.

6 0
3 years ago
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