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Angelina_Jolie [31]
2 years ago
10

What are some other examples of a force acting on an object without touching it?

Physics
1 answer:
a_sh-v [17]2 years ago
8 0
An example would be gravity
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Calculate the kinetic energy of a 5.0kg object moving at 4.0 m/s
Nana76 [90]

Answer:

The answer is

<h2>40 J</h2>

Explanation:

The kinetic energy of an object given it's mass and velocity can be found by using the formula

KE =  \frac{1}{2} m {v}^{2}

where

m is the mass

v is the velocity

From the question

m = 5 kg

v = 4 m/s

The kinetic energy is

KE =  \frac{1}{2}  \times 5 \times  {4}^{2}  \\  =  \frac{1}{2}  \times 5 \times 16 \\  = 5 \times 8

We have the final answer as

<h3>40 J</h3>

Hope this helps you

3 0
3 years ago
What does it mean if the slope is zero?
kiruha [24]
The object is not moving.

My explanation is that say if you sit a ball on the table and it is a smooth surface with no bumps or anything. The ball will sit still since it can’t roll unless you hit it.

Hope this helps!
5 0
3 years ago
For variables control, a circuit voltage will be measured using a sample of five circuits. The past average voltage for samples
astra-53 [7]

Answer:

Average :

UCL = 4.15

LCL = 2.65

Range :

UCL = 2.75

LCL = 0

Explanation:

Given :

Sample size, n = 5

Average, X = 3.4

Range, R = 1.3

A2 for n = 5 ; equals 0.577 ( X chart table)

For the average :

Upper Control Limit (UCL) :

X + A2*R

3.4 + 0.577(1.3) = 4.1501

Lower Control Limit (LCL) :

X - A2*R

3.4 - 0.577(1.3) = 2.6499

FOR the range :

Upper Control Limit (UCL) :

UCL = D4*R

D4 for n = 5 ; equals = 2.114

UCL = 2.114*1.3 = 2.7482

Lower Control Limit (LCL) :

LCL = D3*R

D3 for n = 5 ; equals = 0

LCL = 0 * 1.3 = 0

8 0
3 years ago
"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attach
Lisa [10]

Answer:

<em>The time interval required to lift the spacecraft to this specified height is 123.94 seconds</em>

Explanation:

Height through which the spacecraft is to be lifted = 32.0 m

Mass of the spacecraft = 260.0 kg

Four crew member each pull with a power of 135 W

18.0% of the mechanical energy is lost to friction.

work done in this situation is proportional to the mechanical energy used to move the spacecraft up

work done = (weight of spacecraft) x (the height through which it is lifted)

but the weight of spacecraft = mg

where m is the mass,

and g is acceleration due to gravity 9.81 m/s

weight of spacecraft = 260 x 9.81 = 2550.6 N

work done on the space craft = weight x height

==> work = 2550.6 x 32 = 81619.2 J

this is equal to the mechanical energy delivered to the system

18.0% of this mechanical energy delivered to the pulley is lost to friction.

this means that

0.18 x 81619.2  = <em>14691.456 J  </em> is lost to friction.

Total useful mechanical energy =  81619.2 J - 14691.456 J = 66927.74<em> J</em>

Total power delivered by the crew to do this work = 135 x 4 = 540 W

But we know tat power is the rate at which work is done i.e

P = \frac{w}{t}

where p is the power

where w is the useful work done

t is the time taken to do this work

imputing values, we'll have

540 = 66927.74/t

t = 66927.74/540

time taken t = <em>123.94 seconds</em>

8 0
3 years ago
What happens to electron flow with a conductor of the voltage source is removed?
9966 [12]
The electrons stop flowing
4 0
3 years ago
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