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3 years ago

A car covers 400m distance in 50 sec find it's speed?

Physics

1 answer:

Lena [83]3 years ago

4 0

Answer:

$\boxed {\boxed {\sf 8 \ m/s}}$

Explanation:

Speed is equal to distance over time.

$s=\frac{d}{t}$

The distance is 400 meters and the time is 50 seconds.

$d= 400 \ m \\t= 50 \ s$

Substitute the values into the formula.

$s=\frac { 400 \ m }{ 50 \ s}$

Divide.

$s= 8 \ m/s$

The speed of the car is 8 meters per second.

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irakobra [83]

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4 0

2 years ago

A ray moving in plastic at 62.9 deg enters water, where it bends to 70.9 deg. What is the index of refraction of the plastic?

DedPeter [7]

Answer:

refractive index of plastic is 1.42

Explanation:

When light ray enters from one medium to other medium then due to transition of light it bends away or towards the normal, this phenomenon is known as refraction of light

So here we know that

$n_1 sin i = n_2 sin r$

here we have

$n_2 = \frac{4}{3}$

$i = 62.9^o$

$r = 70.9^o$

now we have

$n_1 sin62.9 = (\frac{4}{3}) sin70.9$

$n_1 = 1.42$

7 0

3 years ago

Calculate Gravitational Potential Energy for an object on Earth with a mass of 2 kg and a height of 7 m.

victus00 [196]

Answer:

137.2J

Explanation:

Ep= mgh

Given,

m=2kg
h=7m

and we know, g = 9.81 N/kg

Ep= 2 × 9.81 × 7

Ep= 137.2J

4 0

3 years ago

A neutron star that spins and emits beams of energy is called a _____.

salantis [7]

Pulsar or also sometimes as a actively rotating accretion disc

5 0

3 years ago

A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.

miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

$\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}$

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

5 0

1 year ago