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konstantin123 [22]
3 years ago
9

If a car stopped then accelerates at 25m/s2 for 2 seconds what is the final velocity of the car?

Physics
1 answer:
nalin [4]3 years ago
5 0
The velcocity equals acceleration times time
which is 50 ms per sec
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A 3.2 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4 + 15.7 x − 1.5 x 2 , where Fx
garik1379 [7]

Answer:

Explanation:

Work: This can be defined as the product of force and distance. The unit of work is Joules (J). it can be expressed mathematically as

W = F×d

or

W = \int\limits^b_a {Fx} \, dx.................................. Equation 1

Where b = upper limit, a = lower limit, Fx = expression of force.

<em>Given: a = 0 , b = 1.3 m, Fx = 4 + 15.7x - 1.5x²</em>

Substituting these values into equation 1

<em>W = \int\limits^a_b {(4 + 15.7x - 1.5x^{2} )dx} \,</em>

W = ᵇ[4x + 15.7x²/2-1.5x³/3 +C]ₐ

Work = upper limit - lower limit

Work = ᵃ[4x + 15.7x²/2 - 1.5x³/3 +C] - [4x + 15.7x²/2 + 1.5x³/3 +C]ᵇ............... Equation 2

Substituting the values of a and b into equation 2

Work = [4(1.3) + 15.7(1.3)²/2-1.5(1.3)³/3 + C] - [0 +C]

Work = [5.2 + 26.53 -3.29 + C] - C

Work = 28.44 J

Work done by the force = 28.44 J.

8 0
3 years ago
Electric power in a circuit is the rate at which _________.
densk [106]
It's the rate that energy is used or supplied
5 0
3 years ago
Explain why it is easier to climb a mountain on a zigzag path rather than one straight up the side.
GaryK [48]
Although a zig zag pattern going up a mountain means you walk further, the incline of the slope is a lot less so you don't have to work as hard.
8 0
3 years ago
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
The magnetic field at point P due to a 2.0-A current flowing in a long, straight, thin wire is 8.0 μT. How far is point P from t
Tanzania [10]

Answer:

r = 0.05 m = 5 cm

Explanation:

Applying ampere's law to the wire, we get:

B = \frac{\mu_oI}{2\pi r}\\\\r =  \frac{\mu_oI}{2\pi B}

where,

r = distance of point P from wire = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = 2 A

B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T

Therefore,

r = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(2\ A)}{2\pi(8\ x\ 10^{-6}\ T)}\\\\

<u>r = 0.05 m = 5 cm</u>

8 0
2 years ago
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