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3 years ago

9

The distance versus time graph for Object A and Object B are shown.

2 answers:

Sergeeva-Olga [200]3 years ago

8 0

Answer:

3

Explanation:

Both move with constant speed.

lianna [129]3 years ago

5 0

Both move with constant speed

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Suppose the total momentum of two masses before a collision is 100 kg m/s. What is the total momentum of the two masses after th

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2 years ago

During the first 14 minutes of a 1.0 hour trip, a car has an average speed 36 km/h. What must the average speed in km/h of the c

oksian1 [2.3K]

Answer:

85.5 km/h

Explanation:

$t_{1}$ = time interval for first phase = 14 min = $\frac{14}{60}$ h = 0.233 h

$t_{2}$ = time interval for second phase = 46 min = $\frac{46}{60}$ h = 0.767 h

$v$ = average speed for the entire trip = 74 km/h

$v_{1}$ = average speed in first phase = 36 km/h

$v_{2}$ = average speed in second phase

$d_{1}$ = distance traveled in first phase

$d_{2}$ = distance traveled in first phase

average speed is given as

$v = \frac{d_{1} + d_{2}}{t_{1} + t_{2}}$

$v = \frac{v_{1} t_{1} + v_{2} t_{2}}{t_{1} + t_{2}}$

$74 = \frac{(36) (0.233) + v_{2} (0.767)}{0.233 + 0.767}$

$74 = (36) (0.233) + v_{2} (0.767)$

$v_{2} = 85.5$ km/h

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3 years ago

4. A woman releases one egg every month for 37 years. Calculate how many

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2 years ago

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

<u>Using the equation of motion:</u>

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

<u>Now the new velocity:</u>

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

3 years ago