What is the total energy released when 9.11 x10^-31 ki?

A force of 110 N is applied at an angle of 30° to the horizontal. Because the force does not align directly either vertically or horizontally with the sled, it can be broken down into two components based on sine and cosine.

For the component parallel to the ground:
x = rcosβ
x = 110cos30°
x = 95.26

For the component perpendicular to the ground:
y = rsinβ
y = 110sin30°
y = 55

3 0

27 days ago

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a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

Keith_Richards [2263]

Answer:

The resulting velocity for him will be 0.187 m/s in reverse direction.

Explanation:

Given:

The mass of the man is, $M=75\ kg$

The mass of the ball is, $m=4\ kg$

The initial velocity of the man is, $u_m=0\ m/s(rest)$

The initial velocity of the ball is, $u_b=0\ m/s(rest)$

The final velocity of the ball is, $v_b=3.50\ m/s$

The final velocity of the man is, $v_m=?\ m/s$

To determine this scenario, we employ the principle of momentum conservation.

This principle states that the total initial momentum equals the total final momentum.

Momentum is calculated by multiplying mass by velocity.

Initial momentum = Initial momentum of the man and the ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of the man and the ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Hence, the total initial momentum equals the total final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign indicates that the man moves backward.

Thus, his final velocity ends up being 0.187 m/s backward.

3 0

4 days ago

A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h

Yuliya22 [2438]

Answer:

Speeds of 1.83 m/s and 6.83 m/s

Explanation:

Based on the law of conservation of momentum,

$mv_o=m(v_1 + v_2)$ where m represents mass, $v_o$ is the initial speed before impact, $v_1$ and $v_2$ are the velocities of the impacted object after the collision and of the originally stationary object after the impact.