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2 years ago

14

How much force is required to accelerate a 50 kg with a mass at 4 m/s2?

1 answer:

GalinKa [24]2 years ago

5 0

Answer:200N

Explanation:

Mass=50kg

Acceleration=4m/s^2

Force=mass x acceleration

Force=50 x 4

Force=200N

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Conductors have ___<br> resistance.

vazorg [7]

Answer:

little/no

Explanation:

Conductors are materials, which conduct electricity and/or heat. That means, that their resistance to such energy is so little, that an electric current is able to pass through.

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3 years ago

Can y'all please help me out with this ?

Inessa05 [86]

Answer:

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2 years ago

A body weighing 108N moves with speed of 5m/s in a horizontal

poizon [28]

Answer:

i) 5 $m$ / $s^{2}$ ii) 54 $N$ iii) 54 $N$

Explanation:

i) $a = \frac{v^{2}}{r}$ ⇒ $a$ = 5² ÷ 5 = 5 $m/s^{2}$

ii) $m = \frac{W}{g}$ ⇒ $m$ = 108 ÷ 10 = 10.8 $kg$ , $F = ma$ ⇒ $F =$ 10.8 × 5 = 54 $N$

iii) F1 = F2 = 54 $N$

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2 years ago

A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of

andre [41]

Answer:

The angle of incidence is greater than the angle of refraction

Explanation:

Refraction occurs when a light wave passes through the boundary between two mediums.

When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where :

$n_1$ is the index of refraction of the 1st medium

$n_2$ is the index of refraction of the 2nd medium

$\theta_1$ is the angle of incidence (the angle between the incident ray and the normal to the boundary)

$\theta_2$ is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

$n_1=1.00$ (index of refraction of air)

$n_2=1.50$ approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

$sin \theta_2 =\frac{n_1}{n_2}sin \theta_1$

And since $n_2>n_1$ , we have

$\frac{n_1}{n_2}$

And so

$\theta_2$

Which means that

The angle of incidence is greater than the angle of refraction

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2 years ago

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

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3 years ago