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1 year ago

A ball is kicked horizontally at 8.0 m/s from a cliff 80m high. What is the acceleration of the ball in the vertical

Physics

1 answer:

blagie [28]1 year ago

6 0

The acceleration of the ball in the vertical direction is $9.8 m/s^2$ , downward.

Explanation:

This is a typical example of projectile motion, which consists of two independent motions:

- A uniform horizontal motion at constant velocity (since there are no forces in the horizontal direction)

- A vertical accelerated motion at constant acceleration (due to the presence of the force of gravity)

We are considering now the vertical motion only. There is only one force acting on the ball in this direction: the force of gravity, of magnitude

F = mg

where m is the mass of the ball and $g=9.8 m/s^2$ the acceleration due to gravity, downward. This means that the acceleration of the ball in this direction is (using Newton's second law)

$a=\frac{F}{m}=\frac{mg}{m}=g = 9.8 m/s^2$

Therefore, the acceleration of the ball in the vertical direction is $9.8 m/s^2$ , downward.

Learn more about projectile motion:

brainly.com/question/8751410

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You might be interested in

A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s east find the vel

Harlamova29_29 [7]

Answer:

38.0 m/s east

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.311) (30.3) + (0.0570) (19.2) = (0.311) v₁ + (0.0570) v₂

10.52 = 0.311 v₁ + 0.0570 v₂

In an elastic collision, kinetic energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(0.311) (30.3)² + (0.0570) (19.2)² = (0.311) v₁² + (0.0570) v₂²

306.5 = 0.311 v₁² + 0.0570 v₂²

Solve the system of equations.

0.311 v₁ = 10.52 − 0.0570 v₂

v₁ = 33.82 − 0.1833 v₂

306.5 = 0.311 (33.82 − 0.1833 v₂)² + 0.0570 v₂²

306.5 = 0.311 (1144 − 12.40 v₂ + 0.03360 v₂²) + 0.0570 v₂²

306.5 = 355.7 − 3.856 v₂ + 0.01045 v₂² + 0.0570 v₂²

0 = 0.06745 v₂² − 3.856 v₂ + 49.16

Use quadratic formula.

v₂ = [ 3.856 ± √(14.87 − 13.26) ] / 0.1349

v₂ = 19.2 or 38.0

We know v₂ isn't 19.2 m/s, so v₂ = 38.0 m/s.

6 0

1 year ago

The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.

abruzzese [7]

Answer:

The equilibrium temperature is

21.97°c

Explanation:

This problem bothers on the heat capacity of materials

Given data

specific heat capacities

copper is Cc =390 J/kg⋅C∘,

aluminun Ca = 900 J/kg⋅C∘,

water Cw = 4186 J/kg⋅C∘.

Mass of substances

Copper Mc = 235g

Aluminum Ma = 135g

Water Mw = 825g

Temperatures

Copper θc = 255°c

Water and aluminum calorimeter θ1= 16°c

Equilibrium temperature θf =?

Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water

McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)

Substituting our data into the expression we have

235*390(255-θf)=

(135*900+825*4186)(θf-16)

91650(255-θf)=(3574950)(θf-16)

23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6

Collecting like terms and rearranging

23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf

8.2*10^6=3.66*10^6θf

θf=80.5*10^6/3.6*10^6

θf =21.97°c

5 0

1 year ago

The suspension cable of a 1,000 kg elevator snaps, sending the elevator moving downward through its shaft. The emergency brakes

tester [92]

Answer:

option (E) 1,000,000 J

Explanation:

Given:

Mass of the suspension cable, m = 1,000 kg

Distance, h = 100 m

Now,

from the work energy theorem

Work done by the gravity = Work done by brake

mgh = Work done by brake

where, g is the acceleration due to the gravity = 10 m/s²

Work done by brake = 1000 × 10 × 100

Work done by brake = 1,000,000 J

this work done is the release of heat in the brakes

Hence, the correct answer is option (E) 1,000,000 J

4 0

2 years ago

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.

myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane. You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

5 0

2 years ago

lidiya [134]

Answer:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

Explanation:

Question

Assuming this question "Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is needed to make your meaning clear. For example, 1*10^5 is not the same as 110^5 . When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol x. For example, for the expression ma,

typing m⋅a would be correct, but mxa would be incorrect".

Solution to the problem

For this case we want to write a expression for ma, and based on the previous info we can write:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

But is not correct do this:

mxa=mxa

axm = mxa

8 0

1 year ago

Read 2 more answers