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Yuri [45]
3 years ago
10

Please help me choose which one the correct answer is.

Physics
1 answer:
earnstyle [38]3 years ago
6 0

Answer:more than g I’m 99.9 % sure of it bc if you just drop the ball versus throwing the ball downward which is faster , throwing the ball downward so more than g hope this helps

Explanation:

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The diagram shows a compressed spring between two carts initially at rest on a horizontal frictionless surface. Cart A has a
EastWind [94]

Answer:

momentum

Explanation:

the impulse is equal for both carts, the momentum for both carts is zero both before the string is cut and after it is cut

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3 years ago
At Magic Mountain there is a ride in which people stand up against the inside wall of a large cylinder of radius 3 m. The cylind
mafiozo [28]

Answer:

0.1308

Explanation:

To keep the rider from sliding down, then the friction force F_f must at least be equal to gravity force F_p

F_f = F_p

\mu N = mg

where μ is the coefficient, N is the normal force acted by the rotating cylinder, m is the mass of a person and g = 9.81 m/s2 is the gravitational acceleration.

According to Newton's 3rd and 2nd laws, the normal force would be equal to the centripetal force F_c, which is the product of centripetal acceleration a_c and object mass m

N = F_c = a_cm

Therefore

\mu a_cm = mg

\mu a_c = g

The centripetal acceleration is the ratio of velocity squared and the radius of rotation

a_c = v^2/r = 15^2 / 3 = 75 m/s^2

Therefore

\mu = g/a_c = 9.81 / 75 = 0.1308

3 0
3 years ago
A motorboat travels 92 km in 2 hours going upstream. It travels 132 km going downstream in the same amount of time. What is the
taurus [48]

Answer:

The speed on boat in still water is  56 \frac{km}{h}  and the rate of the current is   10 \frac{km}{h}

Explanation:

Since speed , v= \frac{Distance\, traveled(D)}{Time\, taken(t)}

Therefore speed of motor boat while traveling upstream is

v_{upstream}=\frac{92}{2}\frac{km}{h}=46\frac{km}{h}

and  speed of motor boat while traveling downstream is

v_{downstream}=\frac{132}{2}\frac{km}{h}=66\frac{km}{h}

Let speed of boat in still water be v_b and rate of current be v_w

Therefore v_{upstream}=v_b-v_w=46\frac{km}{h}   ----(A)

and  v_{downstream}=v_b+v_w=66\frac{km}{h}     ------(B)

Adding equation (A) and (B)  we get

2v_b= (46+66) \frac{km}{h}=112 \frac{km}{h}

=>v_b= 56 \frac{km}{h}   ------(C)

Substituting the value of  v_b in equation (A) we get

v_w= 10 \frac{km}{h}

Thus the speed on boat in still water is  56 \frac{km}{h}  and the rate of the current is   10 \frac{km}{h}

5 0
4 years ago
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