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Alik [6]
3 years ago
13

How much force is needed to accelerate a 1,800kg car at rate of 1.5 m/s2?

Physics
1 answer:
topjm [15]3 years ago
7 0

<u>Given data:</u>

acceleration (a) = 1.5 m/s² ,

           mass (m) = 1800 Kg ,

        Determine F = ?

         <em>From Newtons II law</em>

                          F = m.a   N

                              = 1800× 1.5

                             = 2700 N

<em>2700 N force needed to accelerate the car</em>

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7 0
3 years ago
Read 2 more answers
M (b) Calculate the speed of an electron that is accelerated through the same electric potential difference.
MaRussiya [10]

The speed of a electron that is accelerated from rest through an electric potential difference of 120 V  is 6.49\times10^6m/s

<h3>How to calculate the speed of the electron?</h3>

We know, that the  energy of the system is always conserved.

Using the Law of Conservation of energy,

\triangle K+\triangle U=0

Here, K is the kinetic energy and U is the potential energy.

Now, substituting the formula of U and K, we get:

(\frac{1}{2} mv^2)+(-q\triangle V) =0------(1)

Here,

m is the mass of the electron

v is the speed of the electron

q is the charge on the electron

V is the potential difference

Let v_f and v_i represent the final and initial speed.

Here, v_i =0

Solving for v_f, we get:

v_f=\sqrt{\frac{-2q\triangle V}{m} }

=\sqrt{\frac{2\times1.602\times10^{-19}\times 120}{9.109\times10^{-31}} }

=6.49\times10^6m/s

To learn more about the conservation of energy, refer to:

brainly.com/question/2137260

#SPJ4

4 0
1 year ago
If I travel 300 m east, then 400 m west, what is my distance &amp;<br> displacement?
ad-work [718]

Answer:100m west

Explanation:

6 0
3 years ago
A forklift pushes a box with a force of 500 N across the floor for a distance of 5.0 m, then turns around and pushes with the sa
yanalaym [24]

Answer:

5,000J

Explanation:

Work = Force x Distance

Distance back and forth is canceled out, so either the answer is + or -

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4 0
2 years ago
Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw
Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

\displaystyle{U=\frac{q}{4\pi \epsilon_0r}} (1)

where q is the charge of the particle, \epsilon_0 the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and r is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}

Substituting the values q=1.602 \cdot10^{-19}\ C, \displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}} and r=0.5 \cdot 10^{-9} m we obtain:

\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

6 0
3 years ago
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