What is the speed of the block/bullet system after the collision? express your answer in terms of vi, mw, and mb. hints?

1 answer:

6 0

Having said that,
mV + M0 = (m + M)v; where m = bullet mass, M = block mass, V = bullet velocity upon incoming, and v = the block and bullet velocity (ideal without losses) is the answer you are looking for. Thus, v = V(m/(m + M)) is what you want. m = Mb, M = Mw, and V = Vi.

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Zina [86]

I would say b as well. I’m sorry if it’s wrong

5 0

2 years ago

The radius of Earth is about 6450 km. A 7070 N spacecraft travels away from Earth. What is the weight of the spacecraft at a hei

Triss [41]

Answer:

(a) 1767.43 N

(b) 182.45 N

Explanation:

Radius of earth, R = 6450 km

Weight of person, W = 7070 N

mass of person, m = W / g = 7070 / 9.8 = 721.4 kg

(a) h = 6450 km

The value of acceleration due to gravity on height is given by

$g' = g\left ( \frac{R}{R+h} \right )^2$

$g' = g\left ( \frac{6450}{6450+6450} \right )^2$

g' = g / 4 = 9.8 / 4 = 2.45 m/s^2

The weight of the person at such height is

W' = m x g' = 721.4 x 2.45

W' = 1767.43 N

(b) h = 33700 km

The value of acceleration due to gravity on height is given by

$g' = g\left ( \frac{R}{R+h} \right )^2$

$g' = g\left ( \frac{6450}{6450+33700} \right )^2$

g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2

The weight of the person at such height is

W' = m x g'

W' = 721.4 x 0.253

W' = 182.45 N

3 0

3 years ago

Two paths lead to the top of a big hill. One is steep and direct, while the other is twice as long but less steep. How much more

umka2103 [35]

Answer:

None.

Explanation:

Gravitational potential energy, depends only of the mass on which the gravity is doing work, and the displacement produced by this force.
As displacement depends only on the final and initial positions (in this case, the height of the hill), if we choose as our zero reference level the bottom of the hill, the change in gravitational potential energy will be as follows:

$\Delta U = U_{f} -U_{0} = m*g*h - 0 = m*g*h$