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3 years ago

6

A 10 kg mass rests on a table. What acceleration will be generated when a force of 5 N is applied? a 0.5 m/s2 b 3.5 m/s2 c 5 m/s

2 d 35 m/s2

1 answer:

rosijanka [135]3 years ago

3 0

Answer:

a= 0.5m/s^2

Explanation:

Force applied on an object is known as

F=m.a (Newton's second law states it)

a=F/m

a=5/10=0.5m/s^2

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A projectile is fired at an upward angle of 55° from the top of a 120 m cliff with a speed of 150 m/s. What will be its speed wh

telo118 [61]

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The speed when it strikes the ground below is V= 157.64 m/s < -56.92º .

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V= 150m/s

α= 55º

hi= 120m

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Vx= V * cos α

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clearing t we get the total flying time of the projectile:

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t(fall)= t(total fly) - t(max height)

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3 years ago

A 1300-turn coil of wire that is 2.2 cm in diameter is in a magnetic field that drops from 0.11 T to 0 {\rm T} in 12 ms. The axi

KiRa [710]

Answer:

The induced voltage is $\epsilon = 4.53 \ V$

Explanation:

From the question we are told that

The number of turns is $N = 1300 \ turns$

The diameter is $d = 2.2 \ cm =0.022 \ m$

The initial magnetic field is $B_i = 0.11 \ T$

The final magnetic field is $B_f = 0 \ T$

The time taken is $t = 12 \ ms = 12*10^{-3} \ s$

The radius is mathematically evaluated as

$r = \frac{d}{2}$

substituting values

$r = \frac{0.022}{2}$

$r = 0.011 \ m$

Generally the induced emf is mathematically represented as

$\epsilon = - N * \frac{d\phi}{dt}$

Where $d\phi$ is the change in magnetic flux of the wire which is mathematically represented as

$d \phi = dB* A * cos \theta$

=> $d \phi = (B_f - B_i )* A * cos \theta$

Here $\theta = 0$

since the axis of the coil is parallel to the field

Where A is the cross-sectional area of the coil which is mathematically represented as

$A = \pi * r^2$

$A = 3.142 * 0.011^2$

$A = 3.80*10^{-4} \ m^2$

So the induced emf

$\epsilon = - 1300 * \frac{(0- 0.11) * 3.80*10^{-4}}{12*10^{-3}}$ Here we substituted the values of $d \phi$

$\epsilon = 4.53 \ V$

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4 years ago