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3 years ago

A box with a mass of 12.5kg sits on the floor how high would you need to lift it has a GPE of 568j

Physics

1 answer:

Degger [83]3 years ago

4 0

GPE=mgh
m= 12.5kg
g= 9.81 always
h=?

568=12.5*9.81*h
Solve for h
You will get 4.63m

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# A cheetah can start from rest and attain the velocity 72km/h in 2 seconds. Calculate the acceleration of cheetah

yan [13]

Answer:

Explanation:

Solution,

When a certain object comes in motion from rest, in the case, initial velocity = 0 m/s

Initial velocity ( u ) = 0 m/s

Final velocity ( v ) = 72 km/h ( Given)

We have to convert 72 km /h in m/s

$72 \: km \: per \: hour$

$= 72 \times \frac{1000}{60 \times 60}$

$= 20$ m/s

Final velocity ( v ) = 20 m/s

Time taken ( t ) = 2 seconds

Acceleration (a) = ?

Now,

we have,

$a = \frac{v - u}{t}$

$a = \frac{20 - 0}{2}$

$a = \frac{20}{2}$

$a = 10$ m/s ^2

Hope this helps...

Good luck on your assignment..

7 0

2 years ago

A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q

Kryger [21]

D=s(t) so it would be d=10(.19) d=.19 FOR BITH SNDWERS

3 0

3 years ago

Not sure if it went through last time. Please help asap!

Olin [163]

The equation for force is F=ma. Because we have the value of mass (0.42 kg) and the acceleration (14.8 m/s^2), simply plug them into the equation for force to get
$0.42 \times 14.8 = 6.22$
The answer is 6.22 N because newtons are the unit used to measure force.

8 0

2 years ago

Read 2 more answers

Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o

Anna [14]

The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

Learn more about Tension here brainly.com/question/2287912

#SPJ1

8 0

1 year ago

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.550 m/s. The to

Arisa [49]

Answer: 0.5 m/s

Explanation:

Given

Speed of the sled, v = 0.55 m/s

Total mass, m = 96.5 kg

Mass of the rock, m1 = 0.3 kg

Speed of the rock, v1 = 17.5 m/s

To solve this, we would use the law of conservation of momentum

Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns

When the man throws the rock forward

rock:

m1 = 0.300 kg

V1 = 17.5 m/s, in the same direction of the sled with the man

m2 = 96.5 kg - 0.300 kg = 96.2 kg

v2 = ?

Law of conservation of momentum states that the momentum is equal before and after the throw.

momentum before throw = momentum after throw

53.08 = 0.300 * 17.5 + 96.2 * v2

53.08 = 5.25 + 96.2 * v2

v2 = [53.08 - 5.25 ] / 96.2

v2 = 47.83 / 96.2

v2 = 0.497 ~= 0.50 m/s

3 0

2 years ago