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3 years ago

How do the most common forms of energy work?

Physics

1 answer:

DiKsa [7]3 years ago

7 0

Answer:

activity 2

Explanation:

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A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

3 years ago

A girl throws a ball of mass 0.8 kg against a wall. The ball strikes the wall horizontally with a speed of 11 m/s, and it bounce

Karolina [17]

Answer:

F = 352 N

Explanation:

we know that:

F*t = ΔP

so:

F*t = M $V_f$ -M $V_i$

where F is the force excerted by the wall, t is the time, M the mass of the ball, $V_f$ the final velocity of the ball and $V_i$ the initial velocity.

Replacing values, we get:

F(0.05s) = (0.8 kg)(11m/s)-(0.8 kg)(-11m/s)

solving for F:

F = 352 N

3 0

3 years ago

What number is between 2000 and 2500

Kitty [74]

If its just a number 2,300 is an answer

6 0

3 years ago

At a distance of 0.208 cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485 n/c . wh

Anna35 [415]

We have the equation for electric field E = kQ/ $d^{2}$

Where k is a constant, Q is the charge of source and d is the distance from center.

In this case E is inversely proportional to $d^{2}$

So, $\frac{E_{1} }{E_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}$

$E_{1}$ = 485 N/C

$d_{1}$ = 0.208 cm

$d_{2}$ = 0.620 cm

$E_{2}$ = ?

$\frac{485 }{E_{2}} = \frac{0.620^{2}}{0.208^{2}}$

$E_{2}$ = $\frac{485*0.208^{2}}{0.628^{2}}$

$E_{2}$ = 53.20 N/C

7 0

3 years ago

Read 2 more answers

Formula of relative displacement

Zolol [24]

what is the question?

8 0

3 years ago