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morpeh [17]
2 years ago
11

A golf ball is hit by a club. The graph shows the variation with time of the force exerted on the bal

Physics
2 answers:
sammy [17]2 years ago
6 0

Answer:

D the maximum acceleration of the ball

Explanation:

A the average force on the ball

Total change in momentum of the ball is area under the graph between force and time

So here average force is the ratio of area under the force time graph and total time of the graph

B the change in momentum of the ball

Total change in momentum of the ball is area under the graph between force and time

C the contact time between the ball and the club

Here total time of contact is the total time for which force is acting on the ball due to wall

D the maximum acceleration of the ball

Here since we do not know about the mass of ball we can not find the acceleration.

So here we have correct answer as

D the maximum acceleration of the ball

emmasim [6.3K]2 years ago
5 0
You can't find the acceleration of the ball. The graph tells the force, but you'd also need to know the mass of the ball.
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An open-topped freight car with mass 24,000 kg is coasting without friction along a level track. It is raining very hard, and th
skelet666 [1.2K]

Answer:

(a) v = 3..6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

Explanation:

from the question we have the following:

mass of the car (Mc) = 24,000 kg

initial velocity of the car (u) = 4 m/s

mass of water (Mw) = 3000 kg

final velocity of the car (v) = ?

(a) we can calculate the final momentum of the car by applying the conservation of momentum where

initial momentum = final momentum

Mc x U = (Mc + Mw) x V

24000 x 4 = (24000 + 3000) x v

96,000 = 27000v

v =3.6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

7 0
2 years ago
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
2 years ago
A mover pushes a 255 kg piano
faust18 [17]

Answer:

0.495 ms^{-2}

Explanation:

According to the newton's second law of motion we can apply F=ma hear

Force = mass * acceleration

(assume the piano is moving left side )

←F = ma

F_(pull)+ F_(push)= M*a\\77.5 + 48.7 = 255 *a\\a = 0.495 ms^{-2}

7 0
2 years ago
A car travels 500m in 50s, then 1,500m in 75s. Calculate its averages speed for the whole journey
SIZIF [17.4K]

Answer:

15m/s

Explanation:

500 ÷ 50 = 10m/s

1500 ÷ 75 = 20m/s

10 + 20 = 30

30 ÷ 2 = 15m/s

8 0
2 years ago
Read 2 more answers
A 16-Ω loudspeaker, an 8.0-Ω loudspeaker, and a 4.0-Ω loudspeaker are connected in parallel across the terminals of an amplifier
kolbaska11 [484]

Answer:

2.286 ohm

Explanation:

R1 = 16 ohm

R2 = 8 ohm

R3 = 4 ohm

They all are connected in parallel combination

Let the equivalent resistance is R.

1/R = 1/R1 + 1/R2 + 1/R3

1/R = 1/16 + 1/8 + 1/4

1/R = (1 + 2 + 4) / 16

1/R = 7 / 16

R = 16/7 = 2.286 ohm

6 0
2 years ago
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