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fiasKO [112]
3 years ago
13

A ball at the end of a string is swinging in a horizontal circle of radius 0.85 m. The ball makes exactly 3 revolutions per seco

nd. What is its centripetal acceleration?
Physics
1 answer:
Anettt [7]3 years ago
8 0

Answer:

It's centripetal acceleration is 301.7 m/s²

Explanation:

The formula to be used here is that of the centripetal acceleration which is

ac = rω²

where ac is the centripetal acceleration = ?

ω is the angular velocity = 3 revolutions per second is to be converted to radian per second: 3 × 2π  = 3 × 2 × 3.14 = 18.84 rad/s

r is the radius = 0.85 m

ac = 0.85 × 18.84²

ac = 301.7 m/s²

It's centripetal acceleration is 301.7 m/s²

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You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f
Verdich [7]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

u = 0m/s

Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.

s = ut+\frac{1}{2} at^2

Replacing,

1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2

u =6.4013m/s

The height of the acorn above the meter stick can be calculated as,

v^2 = u^2 +2gh

h = \frac{v^2-u^2}{2g}

h = \frac{6.4013^2-0^2}{2(9.8)}

h = 2.0906m

Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is

h = 2.0906+1.87

h = 3.9606m

4 0
4 years ago
An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The val
OverLord2011 [107]

Answer:

Answer

The Final Quality of teh R-134a in the container  is  0.5056

The Total Heat transfer is Q_{in} = 22.62 KJ

Explanation:

Explanation is  in the following  attachments

3 0
3 years ago
9. Carlos jumps out of a hovering helicopter. After he opens his parachute, his acceleration due to gravity decreases
nata0808 [166]

Answer:

A The force of air resistance becomes equal in magnitude to his weight.

Explanation:

When Carlos jumped out of the helicopter, he fell with a downward force which is proportional to his weight.

His weight is the product of his mass and acceleration due to gravity.

When the parachute is deployed, air resistance between the air molecules and the surface area of the parachute generates a drag force, which acts upwards.

If the drag forces exceeds his weight, then his net force will be directed upwards.

If his weight exceeds the drag forces, then the net force on his body will be downwards.

In the case that his drag force, and his weight are equal, there will be a zero net force on him, and he will experience no acceleration.

So, if after he opens his parachute, his acceleration due to gravity decreases  to zero, then we can say that this is because  the force of air resistance (drag) becomes equal in magnitude to his weight.

5 0
4 years ago
Suppose a skydiver (mass=75kg) is falling toward the earth when the skydiver is 100m above the earth he is moving at 60m/s at th
IRISSAK [1]

Given:

Mass(m)=75kg

Height (h) =100m

v(velocity)=60m/s

a(g)=9.8m/s^2(since it is a free falling object)

Now we know that

v=u+at

We know that

Potential energy=mgh

Where m is the mass

g is the acceleration due to gravity

h is the height above the ground

Substituting the above values we get

Potential energy=75 x 9.8 x 100

=73500N

Now kinetic energy =1/2mv^2

Where m is the mass

v is the velocity

Kinetic energy= 1/2 (75x60 x 60)

Kinetic energy=135000N

Now mechanical energy=

Kinetic energy+ Potential energy

Substituting the values in the above formula we get

Mechanical energy= 73500+135000

=208500N

7 0
3 years ago
Read 2 more answers
#4#4#4vhvyvhvvuvgufygcyfy
Zigmanuir [339]
Unsaturated I'm pretty sure
6 0
4 years ago
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