All categories

3 years ago

A vehicle of mass 100kg has a kinetic energy of 5000 J at an instant. The velocity at that instant is

Physics

2 answers:

snow_lady [41]3 years ago

7 0

Answer:

$\bf\pink{10\:m/s}$

Explanation:

<h2><u>Given</u> :-</h2>

$\sf\red{Mass \ of \ vehicle \ (m) = 100 \ kg}$
$\sf\orange{Kinetic \ energy \ (K.E.) = 5000 \ J}$

$\sf\green{Velocity\: of\: the \:vehicle \:at \:the \:instant}$

<h2><u>Formula to be used</u> :-</h2>

$\sf\blue{K.E. = \dfrac{1}{2} mv^{2}}$

Where,

K.E. = Kinetic energy possessed by the body
M = Mass of the body
V = Velocity of the body

<h2><u>Solution</u> :-</h2>

$\to\:\:\sf\red{K.E. = \dfrac{1}{2}mv^{2}}$

$\to\:\:\sf\orange{5000 = \dfrac{1}{2} \times 100 \times v^{2}}$

$\to\:\:\sf\green{5000 = 50 \times v^{2}}$

$\to\:\:\sf\blue{\dfrac{5000}{50} = v^{2}}$

$\to\:\:\sf\purple{100 = v^{2}}$

$\to\:\:\sf\red{\sqrt{100} = v}$

$\to\:\:\sf\orange{ 10 = v}$

$\to\:\:\bf\pink{v = 10\:m/s}$

Velocity of the vehicle at the instant is $\bf\green{10\:m/s}$

Nikolay [14]3 years ago

5 0

Answer:

23.51 m/s

Explanation:

Parameters given:

Kinetic energy, K = 295 J

Momentum, p = 25.1 kgm/s

a) The kinetic energy of a body is given as:

where m = mass of the body and v = speed of the body

You might be interested in

How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five

Keith_Richards [23]

Answer:

<em>The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = $I$

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = $5I$ (five times larger)

angular speed = ω/5 (five times smaller)

Recall that the kinetic energy of a spinning body is given as

$KE = \frac{1}{2}Iw^{2}$

therefore,

for the first case, the K.E. is given as

$KE = \frac{1}{2}Iw^{2}$

and for the second case, the K.E. is given as

$KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}$

$KE = \frac{1}{10}Iw^{2}$

<em>this is one-tenth the kinetic energy before its spinning characteristics were changed.</em>

<em>This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

6 0

3 years ago

Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.

GalinKa [24]

Wave D has the same wavelength as wave A, but the amplitude is lower. The answer is Wave D.

7 0

3 years ago

I need help with my science.

Ostrovityanka [42]

Answer:

1: 2

2: 5

Explanation:

hope this helps <3

8 0

3 years ago

a skier starts from rest and skis down a 82 meter tall hill labeled h1, into a valley and staught back up another 35 meter hill(

horrorfan [7]

Answer:

She is going at 30.4 m/s at the top of the 35-meter hill.

Explanation:

We can find the velocity of the skier by energy conservation:

$E_{1} = E_{2}$

On the top of the hill 1 (h₁), she has only potential energy since she starts from rest. Now, on the top of the hill 2 (h₂), she has potential energy and kinetic energy.

$mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2}$ (1)

Where:

m: is the mass of the skier

h₁: is the height 1 = 82 m

h₂: is the height 2 = 35 m

g: is the acceleration due to gravity = 9.81 m/s²

v₂: is the speed of the skier at the top of h₂ =?

Now, by solving equation (1) for v₂ we have:

$v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m}$

$v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s$

Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.

I hope it helps you!

6 0

2 years ago

satellite is placed in an elongated elliptical (not circular) orbit around the Earth. At the point in its orbitwhere it is close

GrogVix [38]

Answer:

a) v2=4147.72 m/s

b) stotal=5.53x10^6 m

Explanation:

a) the length from the center of the earth is equal to:

L1=1x10^6+((6.37/2)x10^6)=4.18x10^6 m

the velocity is 5.14 km/s=5.14x10^3 m/s

the farthest distance is equal to:

L2=2x10^6+((6.37/2)x10^6)=5.18x10^6 m

As the angular momentum is conserved, we have to:

I1=I2

m*L1*v1=m*L2*V2, where m is the mass of satelite

clearing v2:

v2=(L1*V1)/L2=(4.18x10^6*5.14x10^3)/5.18x10^6=4147.72 m/s

b) Using the Newton 3rd law:

vf^2=vi^2+2as

where:

a=g=9.8 m/s^2

vf=0

vi=5.14 km/s

s=?

Clearing s:

s=(vf^2-vi^2)/(2g)=((0-(5.14x10^3)^2)/(2*9.8)=1.35x10^6 m

the total distance is equal to:

stotal=s+L1=1.35x10^6+4.18x10^6=5.53x10^6 m

8 0

3 years ago

A vehicle of mass 100kg has a kinetic energy of 5000 J at an instant. The velocity at that instant is​

A vehicle of mass 100kg has a kinetic energy of 5000 J at an instant. The velocity at that instant is