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VashaNatasha [74]
3 years ago
8

Help ;-;

Physics
1 answer:
nasty-shy [4]3 years ago
3 0

Answer:

all qn 1,2,3 have same answer ,. Yes,. hope it helps

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A 300N box on a 43 degree angle.
Ksenya-84 [330]

Answer:is this a question??? I’m so confused

Explanation:

7 0
2 years ago
What is the Gravitational Potential Energy of a 30 kg box lifted 1.5 meters off the ground?
juin [17]

Answer:

<h2>441 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 30 × 9.8 × 1.5

We have the final answer as

<h3>441 J</h3>

Hope this helps you

6 0
3 years ago
An astronaut working with many tools some distance away from a spacecraft is stranded when the "maneuvering unit" malfunctions.
N76 [4]

Answer:

He can return to the spacecraft by sacrificing some of the tools employing the principle of conservation of momentum.

Explanation:

By carefully evaluating his direction back to the ship, the astronaut can throw some of his tools in the opposite direction to that. On throwing those tools of a certain mass, they travel at a certain velocity giving him velocity in the form of recoil in the opposite direction of the velocity of the tools. This is same as a gun and bullet recoil momentum conservation. It is also the principle on which the operational principles of their maneuvering unit is designed.

6 0
3 years ago
Two wheels with fixed hubs and radii 0.51 m and 1.9 m, each having a mass of 3 kg, start from rest. Forces 5 N and F2 are applie
Katarina [22]

Answer:

18.63 N

Explanation:

Assuming that the sum of torques are equal

Στ = Iα

First wheel

Στ = 5 * 0.51 = 3 * (0.51)² * α

On making α subject of formula, we have

α = 2.55 / 0.7803

α = 3.27

If we make the α of each one equal to each other so that

5 / (3 * 0.51) = F2 / (3 * 1.9)

solve for F2 by making F2 the subject of the formula, we have

F2 = (3 * 1.9 * 5) / (3 * 0.51)

F2 = 28.5 / 1.53

F2 = 18.63 N

Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.

3 0
3 years ago
High-speed stroboscopic photographs show that the head of a 210-g golf club is traveling at 56 m/s just before it strikes a 46-g
tamaranim1 [39]

Explanation:

It is given that,

Mass of golf club, m₁ = 210 g = 0.21 kg

Initial velocity of golf club, u₁ = 56 m/s

Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg

After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.

Initial momentum of golf ball, p_i=m_1u_1=0.21\ kg\times 56\ m/s=11.76\ kg-m/s

After the collision, final momentum p_f=0.21\ kg\times 42\ m/s+0.046v

Using the conservation of momentum as :

p_i=p_f

11.76\ kg-m/s=0.21\ kg\times 42\ m/s+0.046v

v = 63.91 m/s

So, the speed of the  golf ball just after impact is 63.91 m/s. Hence, this is the required solution.

3 0
3 years ago
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