All categories

2 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]2 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]2 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

When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it fro

alexandr402 [8]

Answer:

Explanation:

Given

When we drop an object from height , suppose h

it takes time T

using equation of motion

$h=ut+\frac{1}{2}at^2$

where

$h=displacement$

$u=initial\ velocity$

$a=acceleration$

$t=time$

here $u=0$ because it dropped from a certain height

$h=\frac{1}{2}gT^2$

$T=\sqrt{\frac{2h}{g}}$

When height is increases to three times of original height

i.e. $h'=3 h$

then time period becomes

$T'=\sqrt{\frac{2\times 3h}{g}}$

$T'=\sqrt{3}T$

5 0

3 years ago

Which statement BEST explains why the specific heat of water is higher than the specific heat of most other substances?

solniwko [45]

Due to its polarity and hydrogen bonding water can absorb heat without a significant temperature change.. The high specific heat of water helps regulate the rate at which air changes temperature, which is why the temperature change between seasons is gradual instead of sudden, especially near the oceans.

5 0

3 years ago

Read 2 more answers

Help mee pleaseee :)))

Anettt [7]

Answer:

See the explanation below.

Explanation:

Solving the first image question:

C ) The resulting force is defined by Newton's second law which tells us that the sum of the forces on a body is equal to the product of mass by acceleration. That is, there must be a force that acts on a body to produce an acceleration. If there is no acceleration it is because there are no external forces or developed by the body. And if there is no acceleration the body moves at a constant speed, in a straight line, so the response is C.

For the second image, we must remember that weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

W = weight [N]

m = mass [kg]

g = gravity acceleration [m/s²]

Now we have

m = 50 [kg]

ge = Earth gravity acceleration = 10 [m/s²]

gp = Distant planet gravity acceleration = 4 [m/s²]

We = ge*m

We = 10*50 = 500 [N]

Wp =gp*m

Wp = 4*50 = 200 [N]

Therefore the answer is D

For the third image, The mass is always going to be preserved, regardless of where the body or object is in space, its weight is the only one that changes since the gravitational force is modified. That is, the mass on the moon and on Earth will always be the same.

m = 70 [kg]

First, we must calculate the acceleration, by means of the following equation of kinematics.

$v_{f} =v_{o} +a*t$

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 0 (because stars from the rest)

a = acceleration [m/s²]

t = time = 4 [s]

20 = 0 + a*4

20 = 4*a

a = 5 [m/s²]

Now using Newton's second law which tells us that the total force acting on a body is equal to the product of mass by acceleration.

F = m*a

where:

F = force [N] (units of Newtons)

m = mass = 2 [kg]

a = acceleration = 5 [m/s²]

F = 2*5

F = 10 [N]

The body of Figure D, since a total force of 25 [N] to the left acts on it, in the rest of cases the force is zero or much less than 25 [N]

50 + 40 - 35 - 30 = F

F = 25 [N]

8 0

2 years ago

You have two points in a soil. Point A is at 75 cm; point B is at 25 cm above the reference. The capillary potential energy at p

user100 [1]

Answer:

The potential energy at point A is 17.1675 J

Explanation:

The capillary potential is the work expended to bring up a unit mass of liquid to a point in a capillary region from a level liquid surface. It is the capillary potential that facilitates the movement of moisture within soil capillaries

In meteorology it is used to describe the level of saturated soil above the water table

Potential energy is the energy inherent in a body by virtue of its position, therefore the potentials of both point A and B are

Point A, elevation = 75 cm capillary potential = -100 cm

Point B, elevation = 25 cm capillary potential = -200 cm

The total potential energy at point A is

Elevation above reference - capillary potential =75-(-100) = 175 cm

which gives per unit mass

PE = m × g × h = 1 kg × 9.81 m/s ² × 1.75 m = 17.1675 kg·m²/s² = 17.1675 J

8 0

3 years ago

A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on

n200080 [17]

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

6 0

2 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