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yulyashka [42]
3 years ago
8

Net Force (N)

Physics
1 answer:
DaniilM [7]3 years ago
5 0

Explanation:

F = m a

F= 4.0×4.0

F= 16 N

......................................................................................................

F= ma

25= m × 4.998

m= 25/4.998

m= 5.002 kg

......................................................................................................

F=ma

53= 3 × a

a= 53/3

a= 17.666 m/s

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A circular copper loop is placed perpendicular to a uniform magnetic field of 0.50 T. Due to external forces, the area of the lo
pogonyaev

Answer:

6.3\cdot 10^{-4} V

Explanation:

According to Faraday-Newmann-Lenz, the induced emf in the loop is given by:

\epsilon=-\frac{\Delta \Phi}{\Delta t} (1)

where \frac{\Delta \Phi}{\Delta t} is the rate of variation of the magnetic flux through the loop.

We know that the magnetic flux through the loop is given by

\Phi = BA

where B is the magnetic field and A is the area of the loop. Since the magnetic field is constant, we can write the variation of flux as

\Delta \Phi = B \Delta A

So eq.(1) becomes

\epsilon=-B\frac{\Delta A}{\Delta t}

and the problem gives us:

B=0.50 T is the magnetic field

\frac{\Delta \Phi}{\Delta t}=-1.26\cdot 10^{-3} m^2/s is the rate at which the area changes

Substituting into the equation, we find

\epsilon=-(0.50 T)(-1.26\cdot 10^{-3} m^2/s)=6.3\cdot 10^{-4} V=0.63 mV

8 0
3 years ago
A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of
Dominik [7]

Answer:

The centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s²

Explanation:

Given;

distance traveled in the given time = 200 m

time to cover the distance, t = 29.6 s

speed of the runner, v = d / t

v = 200 / 29.6

v = 6.757 m/s

The centripetal acceleration of the runner is given by;

a_c = \frac{V^2}{r}

where;

r is the radius of the circular arc, given as 50 m

Substitute the givens;

a_c = \frac{V^2}{r}\\\\a_c = \frac{(6.757)^2}{50}\\\\a_c = 0.91 \ m/s^2

Therefore, the centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s².

5 0
3 years ago
A projectile is shot from the edge of a vertical cliff 60.0 m above the ocean. It has a speed of 100 m/s and is fired at an angl
Zielflug [23.3K]

Answer:

79.5 m

Explanation:

Let t be the time taken to hit the surface of water and x be the horizontal distance traveled.

use II equation of motion in Y axis direction

h = uy t + 1/2 g t^2

- 60 = - 100 Sin 35 x t - 1/2 x 9.8 x t^2

-60 = - 57.35 t - 4.9 t^2

4.9 t^2 + 57.35 t - 60 = 0

t = \frac{-57.35\pm \sqrt{57.35^{2} + 4 \times 4.9 \times 60}}{2\times 4.9}

By solving we get

t = 0.97 second

The horizontal distance traveled is

x = ux t

x = 100 Cos 35 x 0.97

x = 79.5 m

3 0
3 years ago
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