All categories

2 years ago

A 1,800kg rollercoaster is going down a hill with a kinetic energy of 25,000J. What

Physics

1 answer:

Softa [21]2 years ago

6 0

The velocity of the rollercoaster going down the hill is equal to 5.270 m/s.

<u>Given the following data:</u>

Mass of rollercoaster = 1,800 kilograms.

Kinetic energy =25,000 Joules.

To calculate the velocity of the rollercoaster going down the hill:

<h3>Formula for kinetic energy.</h3>

Mathematically, kinetic energy is calculated by using the formula;

$K.E = \frac{1}{2} MV^2$

<u>Where:</u>

K.E is the kinetic energy.

M is the mass of an object.

V is the velocity of an object.

Making V the subject of formula, we have:

$V =\sqrt{\frac{2K.E}{M} }$

Substituting the given parameters into the formula, we have;

$V =\sqrt{\frac{2\times 25000}{1800} }\\\\V=\sqrt{27.7778}$

Velocity, V = 5.270 m/s

Read more on kinetic energy here: brainly.com/question/17081653

You might be interested in

A crowd in a stadium is heard by a man standing 823.2 m away. The man hears the noise

Reil [10]

Answer: 329.28 m/s

Explanation:

Given that:

Distance of sound = 823.2 m

Time taken for sound = 2.5 seconds

The speed of traveling sound = ?

Speed is obtained by dividing the distance travelled by the time taken for the travel.

Speed = Distance / time

Speed = 823.2m/2.5 seconds

Speed = 329.28 m/s

Thus, the sound traveling as fast as 329.28 m/s

8 0

3 years ago

Two wires of the same material and having the same volume, are fixed

Setler79 [48]

Answer:

48 kg

Explanation:

Given that the two wires are of same material, so their value of young's modulus will be same

Assuming that the wires are cylindrical in shape

As radius of the first wire is half that of the second wire and therefore the area of cross-section of the first wire will be one-fourth of the second wire( ∵ wire is cylindrical, the cross-sectional part will be circle and the area of the circle = π × r² )

As the volume is same for both wires

∴ π × ( $r_{1}$ )² × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$

Here

$r_{1}$ is the radius of the first wire

$r_{2}$ is the radius of the second wire

$l_{1}$ is the length of the first wire

$l_{2}$ is the length of the second wire

⇒ π × (( $r_{2}$ )² ÷ 4) × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$ (∵ radius of first wire is half that of the second wire)

By cancelling the same terms on both sides

we get

$l_{1}$ = 4 × $l_{2}$

⇒ Length of first wire will be four times of the length of second wire

<h3>Strain is defined as the elongation per unit length</h3>

Strain in first wire = ΔL ÷ $l_{1}$ = ΔL ÷ (4 × $l_{2}$ )

where ΔL is the elongation of the wire which in this case is same in both wires

Strain in second wire = ΔL ÷ $l_{2}$

∴ Strain in second wire is four times of strain in first wire

<h3>Stress = F ÷ A</h3>

where F is the force perpendicular to the cross-sectional area

A is the area of cross-section

Force in first wire = $m_{1}$ × g

where $m_{1}$ is the mass hanged to the first wire

g is the acceleration due to gravity

Force in second wire = $m_{2}$ × g

where $m_{2}$ is the mass hanged to the second wire

g is the acceleration due to gravity

Let $A_{1}$ be the cross-sectional area of first wire

$A_{2}$ be the cross-sectional area of second wire

$A_{2}$ = 4 × $A_{1}$ (∵ cross=sectional area of the wire = π × (radius of the wire)² )

Stress in first wire = ( $m_{1}$ × g) ÷ ( $A_{1}$ )

Stress in second wire = ( $m_{2}$ × g) ÷ ( $A_{2}$ ) = ( $m_{2}$ × g) ÷ (4 × $A_{1}$ )

<h3>Young's modulus is defined as Stress per unit strain</h3>

As Young's modulus is same for both wires, Stress per unit strain must be same for both wires

Stress per unit strain of first wire = (( $m_{1}$ × g) ÷ ( $A_{1}$ )) ÷ (ΔL ÷ (4 × $l_{2}$ ))

Stress per unit strain of second wire = (( $m_{2}$ × g) ÷ (4 × $A_{1}$ )) ÷ (ΔL ÷ $l_{2}$ )

By equating them we get

$m_{2}$ = 16 × $m_{1}$

⇒ $m_{2}$ = 16 × 3 = 48 kg

∴ $m_{2}$ = 48 kg

5 0

3 years ago

PLSSS HELPPPPP!! due today @ 8:30!

romanna [79]

Answer:

(1) weight = 565N

distance = 325m

time= 12.6s

power = work done/time

work done= force * distance

= 565*3.25 =1836.25

:. power = 1836.25\12.6

=145.7!

(2) the books will add more to her weight and thereby increasing her power output.

4 0

3 years ago

A 41-kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.5 m/s. Neglect air resistan

postnew [5]

The altitude of the pole vaulter as she crosses the bar is 5 m.

<h3>The altitude of the bar</h3>

v² = u² - 2gh

where;

v is final velocity of the pole vaulter
u is the initial velocity of the pole vaulter
h is altitude of the bar

h = (u² - v²)/2g

h = (10² - 1.5²)/(2 x 9.8)

h = 5 m

Thus, the altitude of the pole vaulter as she crosses the bar is 5 m.

Learn more altitude here: brainly.com/question/1159693

#SPJ1

3 0

2 years ago

How does the space shuttle maneuver in exosphere

Gemiola [76]

<span>In a rocket combustion chamber the rocket exhaust is forced out of the nozzle by the rapid expansion of exhaust gases and the directing shape of the rocket nozzle. The equal and opposite reaction is that the rocket nozzle (and hence the shuttle it is attached to) is forced in the other direction</span>

8 0

3 years ago