I believe the answer is B.
<h3>Reducing Surface Area.</h3>
If the surface area becomes smaller, the pressure becomes larger.
Answer:
15 m/s or 1500 cm/s
Explanation:
Given that
Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s
Distance moved during the hook, d(h) = 5 cm = 0.05 m
Distance moved by the fist, d(f) = 100 cm = 1 m
Average speed of the fist during the hook, v(f) = ? cm/s = m/s
This can be solved by a very simple relation.
d(f) / d(h) = v(f) / v(h)
v(f) = [d(f) * v(h)] / d(h)
v(f) = (1 * 0.75) / 0.05
v(f) = 0.75 / 0.05
v(f) = 15 m/s
Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s
Answer:
24.084 m/s
Explanation:
From the law of conservation of linear momentum
Total momentum before collision equals to the total momentum after collision
Since momentum=mv where m is mass and v is velocity
where 
is the mass of the truck, 
is velocity of the truck, 
is the common velocity of moving and standing truck after collision and 
is the mass of the standing truck
Making the subject we obtain
Substituting as 25000 Kg, as 22.3 m/s, as 2000 Kg we obtain
Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive
The truck was moving at 24.084 m/s
Hello,
<span>A car with a mass of 2.0×10^3 kg is traveling at 15m/s. We need to find the momentum of the car. To do so, follow this formula:
p=mv
Where,
p = momentum
m = mass
v = </span>velocity
The cars mass is 2.0E3 and its velocity is 15m/s. Therefore:
p=2.0 x 10^3 *15 or 2000(15)
p=30000
Thus, the cars momentum is 30000 kg m/s
Faith xoxo
