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frozen [14]
3 years ago
7

A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory (ignorin

g ar resistance)? a)There is none b)Gravity c)A normal force d)Thrust
Physics
1 answer:
Nat2105 [25]3 years ago
8 0

Answer:

b) Gravity

Explanation:

Gravity acts all of the time, when you apply force to a projectile it has to be more than the forces of the gravity and air resistance together so the projectile can move,  when the rock is at the top of its trajectory the force that you applied at the beginning is getting lost, so the other forces (air resistance and gravity) make the rock fall to the floor.

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12.5 Round to the nearest hundredth.
7nadin3 [17]
I think its 13...........
7 0
3 years ago
While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. how much time does it ta
Yuri [45]

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. The time taken by the car will be 4 seconds. the correct answer is option(c).

When acceleration is constant, the rate of change in velocity is also constant. In the absence of any acceleration, velocity remains constant. When acceleration is positive, velocity becomes more significant.

Let a denote acceleration, u denote initial velocity, v denote final velocity, and t denote time.

The equation of motion is stated as,

v = u + at

v² = u² + 2as

A car travels across a distance of 60.0 m while accelerating constantly from 12.0 m/s to 18.0 m/s.

Then the time taken by the car will be,

u = 12 m/s

v = 18 m/s

s = 60 m

Put these in the equation v² = u² + 2as.

18² = 12² + 2 x a x 60

a = 1.5

Then the time will be

18 = 12 + 1.5t

1.5t = 6

t = 4 seconds

Hence, the time taken is 4 s.

The complete question is:

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. How much time does it take?

1.00 s

2.50 s

4.00 s

4.50 s

To know more about acceleration refer to:  brainly.com/question/12550364

#SPJ4

8 0
2 years ago
two boxes sit on a frictionless surface and are in contact with one another. the first box has a mass of 7 kg and the second box
egoroff_w [7]
The acceleration of the boxes depends on the mass and weight. 

we have a mass of 7 and 8 kilograms

if it took 25 N force to move box A, then you would take 25 and multiply by 8 then divide by 2. 

It will leave you with 100 N. 

finally take the sq rt of 100 to get 10
7 0
3 years ago
At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai
zhannawk [14.2K]

Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

<u>Motion of 0.50 kg ball:</u>

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

<u>Motion of 0.25 kg ball:</u>

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          \bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         \bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m

The center of mass of the two-ball system is 7.05 m above ground.  

8 0
3 years ago
How many significant digits are measurement 0.00210 mg?
jek_recluse [69]
There are 3 significant figures, if that answers the question.
8 0
3 years ago
Read 2 more answers
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