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vichka [17]
3 years ago
8

If a net vertical force of 14.7 N is applied to a ball of mass 1.5 kg what acceleration

Physics
1 answer:
vfiekz [6]3 years ago
6 0

Answer:

9.8m/s^2

Explanation:

F=ma

F=14.7N

m=1.5kg

a=F/m

a=14.7N/1.5kg

a=9.8m/s^2

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One day in second grade, Chris drank chocolate milk before recess. She then played hard during recess and got sick from the heat
rosijanka [135]

Answer:

its A i took the test

Explanation:

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3 years ago
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At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81m/s2 . A watermelon has a weight of 58.0N at the sur
kvasek [131]

Answer: A) mass on earth surface = 5.91kg

B) mass on surface of jupiter = 5.91kg

C) weight on surface of jupiter = 10.697N

Explanation:

The relationship between weight (W), mass (m) and acceleration due gravity (g) is given below

W=mg

From the question, g= 9.8m/s² and weight on the surface on the earth is 58N

A) The mass of watermelon on earth is

m = 58/ 9.8 = 5.91kg

B) the mass of the watermelon on jupiter is 5.91kg.

You will notice this is the same as the mass of watermelon on earth and that is so because mass is a scalar quantity that does not depends on the distance away from the center of the earth (unlike weight which is a vector) thus making it constant all through any location.

C) mass of watermelon is 5.91kg, g=9.8m/s² weight of watermelon on jupiter is given below as

W = mg

W = 5.91 x 9.8

= 10.697N.

6 0
3 years ago
What do radio waves and microwaves have in common?
Tamiku [17]

Answer:

I Will say the Answer is A

Explanation:

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3 years ago
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If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for
steposvetlana [31]

Answer:

\ d_{out} = 100 \ m.

Explanation:

Given data:

F_{in} = 50 \ \rm N

F_{out} = 10 \ \rm N

d_{in} = 20 \ m

Let the distance traveled by the object in the second case be d_{out}.

In the given problem, work done by the forces are same in both the cases.

Thus,

W_{in} = W_{out}

F_{in}.d_{in} = F_{out}.d_{out}

\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}

\ d_{out} = \frac{50 \times 20}{10}

\ d_{out} = 100 \ m.

5 0
3 years ago
A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with
Veseljchak [2.6K]

Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

Starting point. Higher

            Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

            Em_{f} = K = ½ m v²

            Em₀ = Em_{f}

            ½ m v₀² + m g h = ½ m v²

            v² = v₀² + 2 g h

             

Let's calculate

             v = √(1.23² + 2 9.8 1.69)

             v = 5.89 m / s

In the horizontal part we can use the relationship between work and the variation of kinetic energy

            W = ΔK

            -fr x = 0- ½ m v²  

               

Newton's second law

              N- W = 0

     

The equation for the friction is

               fr = μ N

               fr = μ m g

We replace

             μ m g x = ½ m v²

             x = v² / 2μ g

Let's calculate

            x = 5.89² / (2 0.255 9.8)

            x = 6.94 m

6 0
4 years ago
Read 2 more answers
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