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4 years ago

The rate at which an object’s velocity changes is called its

Physics

2 answers:

ANTONII [103]4 years ago

7 0

Answer:

Acceleration

Explanation:

topjm [15]4 years ago

6 0

I believe it is call “Acceleration”

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From a window that is 20 m from the ground a stone with a speed of 10m / s is thrown vertically upwards. Calculate:

Oduvanchick [21]

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the maximum height

v = final velocity at the maximum height = 0 m/s

t = time taken to reach the maximum height

Using the equation

v² = v₀² + 2 a (Y - Y₀)

0² = 10² + 2 (- 9.8) (Y - 20)

Y = 25.1 m

also using the equation

v = v₀ + a t

inserting the values

0 = 10 + (- 9.8) t

t = 1.02 sec

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the ground = 0 m

t = time taken to reach the ground

Using the equation

Y = Y₀ + v₀ t + (0.5) a t²

0 = 20 + 10 t + (0.5) (- 9.8) t²

t = 3.3 sec

3 0

3 years ago

When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do

Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force

$\Delta p$ is the change in momentum

$\Delta t=0.030 s$ is the time interval

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

$v=281 km/h =78.1 m/s$ is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N$

7 0

3 years ago

When you jump, you exert a pushing force against the ground. Gravity pulls you back down. Why can a person jump higher on the mo

Dennis_Churaev [7]

Answer:

This is because the force of gravity is much less on the moon than on the earth, therefore the person wont be pulled down much and will jump higher

7 0

3 years ago

When the core of a star like the sun uses up its supply of hydrogen for fusion, the core begins to ________?

pav-90 [236]

When the core of a star like the sun uses up its supply of hydrogen for fusion, the core begins to contract in size. The gravitational forces becomes strong and cause the star to reduce in size. as the star contracts, the temperature of the star also rises since the heat is distributed inside smaller Volume now.

4 0

4 years ago

Read 2 more answers

Let’s tie this all together. It makes sense that, if the rope force remains greater than the gravitational force, the child keep

Pavlova-9 [17]

Answer:

If child weight is equal to rope force then child will move with uniform speed

or we can say that the child will remain at rest in his position

Explanation:

As we know that child is hanging by rope

so here there will be two forces on the child

1) Weight or gravitational force which act vertically downwards

2) Tension in the rope which act vertically upwards

Now if child will accelerate upwards then tension force must be more than the weight of the child

If tension force is less than the weight then child will decelerate and his speed will decrease

if tension force is equal to child weight then in that case the child will remain at rest or it will move with same speed

8 0

3 years ago