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3 years ago

What is one force listed below which helps to stop the

Physics

2 answers:

oksian1 [2.3K]3 years ago

8 0

Answer:

applied force

Explanation:

An applied force is a force that is applied to an object by a person or another object. If a person is pushing a desk across the room, then there is an applied force

V125BC [204]3 years ago

8 0

The answer is b hope it helped

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Which type of power plants generate the least amount of pollution? A) Biomass power plants B) Geothermal power plants C) Nuclear

sladkih [1.3K]

Answer:

Explanation:

6 0

3 years ago

An electron moves through a uniform electric field E = (2.00î + 5.40ĵ) V/m and a uniform magnetic field B = 0.400k T. Determine

EleoNora [17]

Answer: $a=1.75\times 10^{21}\left ( 2\hat{i}5.08\hat{j}\right )$

Explanation:

Given

Electric Field $\vec{E}=2\hat{i}+5.4\hat{j}$

$\vec{B}=0.4\hat{k}\ T$

velocity $\vec{v}=8\hat{i} m/s$

mass of electron $m=9.1\times 10^{-31} kg$

Force on a charge Particle moving in Magnetic Field

$F=e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]$

$a=\frac{e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]}{m}$

$a=\frac{1.6\times 10^{-19}\left [ 2\hat{i}+5.4\hat{j}+\left ( 8\hat{i}\times 0.4\hat{k}\right )\right ]}{9.1\times 10^{-31}}$

$a=1.75\times 10^{21}\left ( 2\hat{i}+5.08\hat{j}\right )\ m/s^2$

7 0

3 years ago

The strength of a magnetic field around a wire carrying a current of 20 A is 0.004 T. What is the strength of the

Alex777 [14]

Answer:

D. 0.008 T

Explanation:

Eng 2021

7 0

3 years ago

If the car passes point A with a speed of 20 m/s and begins to increase its speed at a constant rate of at = 0.5 m/s2 , determin

IceJOKER [234]

Answer:

$1.68 \frac{m}{s^2}$

Explanation:

Please find the image for the question as attached file.

Solution -

Given -

First of all we will calculate the velocity at point C,

As per newton's third law of motion-

$V_C^2 = V_A^2 + 2 a_t (S_C - S_A)\\$

Substituting the given values in above equation, we get -

$V_C^2 = 20^2 + 2*0.5*(100-0)\\V_C = 22.361 \frac{m}{s}$

Now we will determine the radius of curvature for the curve shown in the attached image

$Y = 16 - \frac{1}{625} X^2\\$

Differentiating on both the sides, we get -

$\frac{dy}{dx} = -3.2 (10^-3) X\\\frac{d^2y}{d^2x} = -3.2 (10^-3)\\Curve = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}} }{\frac{d^2y}{d^2x}} \\Curve = 312.5$ meter