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BlackZzzverrR [31]
3 years ago
10

You have a bowling ball with a mass of 4kg. You throw it with an acceleration of 10 m/s/s. With how much force will it hit the p

ins?
Physics
1 answer:
Eddi Din [679]3 years ago
6 0

Answer:

<h3>The answer is 40 N</h3>

Explanation:

The force acting on an object can be found by using the formula

<h3>force = mass × acceleration</h3>

From the question

mass = 4 kg

acceleration = 10 m/s²

So we have

force = 4 × 10

We have the final answer as

<h3>40 N</h3>

Hope this helps you

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Which type of wave is not a light wave
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4 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
During the 28-day lunar cycle, the positions of the Sun,
Mnenie [13.5K]
The answer is full moon because that’s how it changes in life
8 0
3 years ago
A 63 gg ice cube can slide without friction up and down a 30∘30∘ slope. The ice cube is pressed against a spring at the bottom o
tatiyna

Given Information:

slope angle = θ = 30°

spring constant = k = 30 N/m

compressed length = x = 10 cm = 0.10 m

mass of ice cube = m = 63 g = 0.063 kg

Required Information:

distance traveled by ice cube = d = ?  

Answer:

distance traveled by ice cube = 0.48 m

Explanation:

Using the the principle of conversation of energy, the following relation holds true for this case,

mgh = 1/2*kx²

h = 1/2*kx²/mg

Where h is the height of the slope, m is the mass of ice cube, k is the spring constant and x is the compressed length o the spring and g is gravitational acceleration.

h = 1/2*kx²/mg

h = 1/2*30(0.1)²/0.063*9.8

h = 0.242 m

From trigonometry ratio,

sinθ = h/d

d = h/sinθ

d = 0.242/sin(30)

d = 0.48 m

Therefore, when the ice cube is released, it will travel a total distance 0.48 up the slope before reversing direction.

3 0
3 years ago
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