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2 years ago

9

a truck starts to move from rest if it given the acceleration of 3 metre per second square in 5 second calculate the final veloc

ity and distance covered by truck in full process

1 answer:

frosja888 [35]2 years ago

7 0

Hope this helps. Plzz let me know if you have any further doubts :)

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A light ray incident on a block of glass makes an incident angle of 50.0° with the normal to the surface. The refracted ray in t

Andru [333]

Answer:

$n_{glass} = 1.65$

Explanation:

As we know that the angle of incidence is given as

$i = 50.0^o$

also we have angle of refraction as

$r = 27.7^o$

now by Snell's law we know that

$n_{air} sin i = n_{glass} sin r$

$1 sin50 = n_{glass} sin 27.7$

now we have

$n_{glass} = \frac{sin 50}{sin 27.7}$

$n_{glass} = 1.65$

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3 years ago

A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be

MrRissso [65]

The kinetic energy halfway the hill is $2.86\cdot 10^5 J$

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

$U_A +K_A = U_B + K_B$

where

$U_A=mgh_A$ is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

$g=9.8 m/s^2$ (acceleration of gravity)

$h_A = 30 m$ (height at point A)

$K_A=\frac{1}{2}mv_A^2$ is the initial kinetic energy, at point A , with

$v_A=17 m/s$ (velocity at point A)

$U_B=mgh_B$ is the final potential energy, at point B, where

$h_B = 15 m$ (height at point B)

$K_B=\frac{1}{2}mv_B^2$ is the final kinetic energy, at point B, where

$v_B$ is the velocity at point B

Here we are interested in finding $K_B$ , so by re-arranging the equation and substituting we find:

$K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0

3 years ago

A ball is dropped from a window and takes two seconds to reach the ground .it starts from rest and reaches a final speed of 20m/

Akimi4 [234]

The ball can't reach the speed of 20 m/s in two seconds, unless you THROW it down from the window with a little bit of initial speed. If you just drop it, then the highest speed it can have after two seconds is 19.6 m/s .

If an object starts from rest and its speed after 2 seconds is 20 m/s, then its acceleration is 20/2 = 10 m/s^2 .

(Gravity on Earth is only 9.8 m/s^2.)

3 0

3 years ago

The amplitude of a paricular wave is 4.0 m. The crest to trough distance

kozerog [31]

Answer:

The crest to trough distance = 8 m

Explanation:

Given that,

The amplitude of a particular wave is 4.0 m.

We need to find the crest to trough distance.

We know that,

Amplitude = The distance from the base line to the crest or the the distance from the baseline to the trough.

It means,

Distance from crest to trough = 2(Amplitude)

= 2(4)

= 8 m

Hence, the crest to trough distance is equal to 8 m.

6 0

2 years ago

Why do we see black? Is light being admitted reflected or absorbed or transmitted

Diano4ka-milaya [45]

Answer:When white light strikes an object, each individual frequency of light is transmitted, reflected, or absorbed, depending on the properties of the surface molecules. If all frequencies are absorbed by the object, then it appears black. If all frequencies are reflected, then it appears white.

Explanation:

4 0

2 years ago