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gtnhenbr [62]
3 years ago
14

What force is needed to accelerate an object 5 m/s squared if the object has a mass of 10 kilograms

Physics
1 answer:
Alborosie3 years ago
5 0
50N......................
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What is the net force required to accelerate a 884 kg car at 2 m/sec2?
EastWind [94]

Answer:

1768 N

Explanation:

We can solve the problem by using Newton's second law:

F=ma

where

F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, we have a car of mass

m = 884 kg

And its acceleration is

a=2 m/s^2

Substituting into the equation, we find the net force on the car:

F=(884)(2)=1768 N

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How to solve this question on reflection
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3 years ago
A backpack has a mass of 8 kg. It is lifted and given 54.9 J of gravitational potential energy. How high is it lifted? Accelerat
sweet [91]
Potencial Energy=hma
Where
Potencial Energy =E= 54.9J
h=?
m=8kg
a=9.8m/s^2
You need to know that 1 J=1(kgm^2)/s^2

Isolate h=E/(ma)
h=(54.9)/(8*9.8)
8 0
3 years ago
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A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential
ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is 1.022\times10^{-16}\ J

(c). The work done on the proton is -8\times10^{-18}\ J.

Explanation:

Given that,

Speed v= 3.50\times10^{5}\ m/s

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

U=qV=eV

Using conservation of energy

K_{i}+U_{i}=K_{f}+U_{f}

\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}

]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})

\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V

\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}

Put the value into the formula

\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}

\Delta V=639.2=0.639\ kV

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

\Delta U=q\Delta V

Put the value into the formula

\Delta U=1.6\times10^{-19}\times639.2

\Delta U=1.022\times10^{-16}\ J

(c). We need to calculate the work done on the proton

Using formula of work done

\Delta U=-W

W=q(V_{2}-V_{1})

W=-1.6\times10^{-19}(150-100)

W=-8\times10^{-18}\ J

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is 1.022\times10^{-16}\ J

(c). The work done on the proton is -8\times10^{-18}\ J.

5 0
4 years ago
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