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2 years ago

7

A spring is stretched 2.0 cm. If the same spring is stretched 8.0 cm the ratio of the second and first potential energies of the

spring is
16
9
4
1

1 answer:

jok3333 [9.3K]2 years ago

6 0

Potential energy of spring equals K times X squared divided by 2 where X is displacement

4 times squared equals 16

choose 1st answer

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Calculate the acceleration of a 1400-kg car that stops from 39 km/h "on a dime" (on a distance of 1.7 cm).

Vinil7 [7]

Answer:

$a=-3449.67\frac{m}{s^2}$

Explanation:

The car is under an uniforly accelerated motion. So, we use the kinematic equations. We calculate the acceleration from the following equation:

$v_f^2=v_0^2+2ax$

We convert the initial speed to $\frac{m}{s}$

$39\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=10.83\frac{m}{s}$

The car stops, so its final speed is zero. Solving for a:

$a=\frac{v_0^2}{2x}\\a=-\frac{(10.83\frac{m}{s})^2}{2(1.7*10^{-2}m)}\\a=-3449.67\frac{m}{s^2}$

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3 years ago

)If a force of 5.00 N is needed to open a 90.0 cm wide door when applied at the edge opposite the hinges, what force must be app

masha68 [24]

Answer:

A force of 12.857 newtons must be applied to open the door.

Explanation:

In this case, a force is exerted on the door, a moment is performed and the door is opened. If moment remains constant, the force is inversely proportional to distance respect to axis of rotation passing through hinges. That is:

$F \propto \frac{1}{r}$

$F = \frac{k}{r}$ (Eq. 1)

Where:

$F$ - Force, measured in newtons.

$k$ - Proportionality ratio, measured in newton-meters.

$r$ - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

$F_{A}\cdot r_{A} = F_{B}\cdot r_{B}$

$F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}$ (Eq. 2)

Where:

$F_{A}$ , $F_{B}$ - Initial and final forces, measured in newtons.

$r_{A}$ , $r_{B}$ - Initial and final distances, measured in meters.

If we know that $F_{A} = 5\,N$ , $r_{A} = 0.9\,m$ and $r_{B} = 0.35\,m$ , then final force is:

$F_{B}= \left(\frac{0.9\,m}{0.35\,m} \right)\cdot (5\,N)$

$F_{B} = 12.857\,N$

A force of 12.857 newtons must be applied to open the door.

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2 years ago

What is the second law of thermodynamics

andriy [413]

It states that the total entropy of an isolated system can never decrease over time

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2 years ago

Which law of motion explains what happens during a ride on the bumper cars ?

fomenos

Answer:

Newtows first law of motion

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1 year ago

A car accelerates uniformly from rest to speed 6.6 m/s in 6.5 s .Find the distance the car travel during this time .

kirill [66]

Answer:

<em>The distance the car traveled is 21.45 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Solving [1] for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

$\displaystyle a=\frac{6.6-0}{6.5}$

$a = 1.015\ m/s^2$

The distance is now calculated with [2]:

$\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}$

x = 21.45 m

The distance the car traveled is 21.45 m

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2 years ago