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3 years ago

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The mass of car is 1200kg. If the car is moving with tue velocity of 30m/s. How much force should be applied to stop the car in

20 seconds?

1 answer:

faust18 [17]3 years ago

7 0

Answer:

momentum of car=1200kg *10m/s =12000kgm/s

momentum =force * time

12000 kgm/s =force * 20s

force = 600 N

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A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60∘ away from vertical increases fro

bixtya [17]

Answer:

The induced emf is $|\epsilon|=0.0261 V$

Explanation:

From the question we are told that

The number of turn is $N = 100$

The diameter of the coil is $d = 2.0 cm = 2.0 *10^{-2} m$

The uniform magnetic at initial is $B_i = 0.50 T$

The uniform magnetic at initial is $B_f = 1.50 T$

The time taken is $t = 0.60s$

The angle the magnetic field makes with vertical is $\theta = 60^o$

Generally induced emf is mathematically represented as

$\epsilon = -N \frac{d \o}{dt}$

where $d \o$ is the change magnetic flux

Magnetic flux is mathematically represented as

$\O = \= B \cdot \= A$

$= BA cos \theta$

Substituting this above

$\epsilon = -N \frac{d (BA cos \theta)}{dt}$

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

Where B is the magnetic field and A is the area which is mathematically evaluated as

$A = \frac{\pi d^2}{4}$

Substituting values

$A = \frac{\pi (2.0 *10^{-2})^2}{4}$

$A= 3.142*10^{-4}m^2$

From the equation of emf

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

dB = $B_2 -B_1$

So

$|\epsilon| = N A \frac{ (B_2 -B_1 cos \theta)}{dt}$

substituting values

$|\epsilon| = 100(3.142*10^{-4}) \frac{ (1.50 -0.50 cos(60))}{0.60}$

$|\epsilon|=0.0261 V$

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Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

lina2011 [118]

Answer:

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Explanation:

To find the x- and y- components, we have to apply the formulas:

$v_x = v cos \theta$

$v_y = v sin \theta$

where

v = 12.0 m/s is the magnitude of the vector

$\theta$ is the angle between the direction of the vector and the positive x-axis

Here, the angle given is the angle above the negative x-axis; this means that the angle with respect to the positive x-axis is

$\theta=180^{\circ} - 60^{\circ} = 120^{\circ}$

So, the two components are:

$v_x = (12.0 m/s) cos 120^{\circ}=-6.0 m/s$

$v_y = (12.0 m/s) sin 120^{\circ}=10.4 m/s$

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