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3 years ago

A 1000 kg car drives off of u 30m cliff of

Physics

1 answer:

Inessa05 [86]3 years ago

5 0

The mechanical energy of the car is $1.54\cdot 10^6 J$

Explanation:

The mechanical energy of an object is equal to the sum of its gravitational potential energy (GPE) and its kinetic energy (KE):

E = GPE + KE

The GPE of the car is given by:

$GPE = mgh$

where

m = 1000 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

h = 30 m is the height of the car above the ground

Substituting,

$GPE=(1000)(9.8)(30)=2.94\cdot 10^5 J$

The kinetic energy of the car is given by:

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

where

m = 1000 kg

v = 50 m/s is the speed of the car

Substituting,

$KE=\frac{1}{2}(1000)(50)^2=1.25\cdot 10^6 J$

Therefore, the mechanical energy is

$E=GPE+KE=2.94\cdot 10^5 + 1.25\cdot 10^6 = 1.54\cdot 10^6 J$

Learn more about potential energy and kinetic energy:

brainly.com/question/6536722

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Sort the processes based on the type of energy transfer they involve.

alina1380 [7]

Types of energy transfer involve
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4 years ago

The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick

Travka [436]

Answer:

c. vf is greator than v2, but less than v1

Explanation:

The principle of conservation of linear momentum states that when two or more bodies act upon one another, their total momentum remains constant.

In a system of colliding bodies the total momentum of the system just before the collision is the same as the total momentum just after the collision.

Collisions in which the kinetic energy is conserved are called elastic collision.

Collisions in which the kinetic energy is not conserved are called inelastic collisions. If the two objects stick together after the collision and move with a common velocity, the collision is said to be perfectly inelastic.

<em>The above scenario is a perfectly inelastic collision. The initial velocity of particle 1 was greater than particle 2 before collision. After collision, its velocity will reduce to a final velocity vf as it transfers some of its kinetic energy to particle 2; whereas, the velocity of particle 2 will increase to a final velocity vf as it absorbs some of the kinetic energy of particle 1.</em>

Therefore,

a. vf = v2 is wrong because vf is greater than v2

b. vf is less than v2 is wrong because vf is greater than v2

c. vf is greater than v2, but less than v1 is correct.

d. vf = v1 is wrong because vf is less than v1

4 0

3 years ago

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the ot

s344n2d4d5 [400]

Answer:

The length of the longer pipe is L = 2.30 m

Explanation:

Given that:

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.

How long is the other pipe?

From above;

The formula for the frequency of open ended pipes can be expressed as:

$f = \dfrac{nv}{2L}$

where n = 1 ( since half wavelength exist between those two pipes)

v = 343 m/s and L = 2.08 m

Thus, the shorter pipe produces a frequency of :

$f = \dfrac{1*343}{2*2.08}$

$f = \dfrac{343}{4.16}$

$f =82.45 \ Hz$

Also; we know that the beat frequency was given as 8.0 Hz

Then,

The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz

The lower frequency of the longer pipe = 74.45 Hz

Finally;

From the above equation; make Length L the subject of the formula. Then,

The length of the longer pipe is L = $\dfrac{nv}{2f}$

The length of the longer pipe is L = $\dfrac{1*343}{2*74.45}$

The length of the longer pipe is L = $\dfrac{343}{148.9}$

The length of the longer pipe is L = 2.30 m

6 0

3 years ago

When you push a 2.00 kg book resting on a tabletop it takes 4.60 N to start the book sliding. What is the coefficient of static

natali 33 [55]

The coefficient of static friction is 0.234.

Answer:

Explanation:

Frictional force is equal to the product of coefficient of friction and normal force acting on any object.

So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.

Normal force = mass * acceleration due to gravity

Normal force = 2 * 9.8 = 19.6 N.

And the frictional force is given as 4.6 N, then

$Coefficient of static friction = Frictional force/Normal force$

Coefficient of static friction = 4.6 N / 19.6 N = 0.234

So the coefficient of static friction is 0.234.

3 0

3 years ago

A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane

Mama L [17]

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, $\phi=8.6\times 10^{-3}\ T/m^2$

The magnetic flux linked through the loop is :

$\phi=B.A$

$\phi=BA\ cos\theta$

Here, $\theta=0$

$B=\dfrac{\phi}{A}$

$B=\dfrac{\phi}{\pi r^2}$

$B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}$

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

4 0

3 years ago