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3 years ago

How does the speed of an object relate to the energy of the object

Physics

1 answer:

Ann [662]3 years ago

7 0

Answer:

Increasing the speed of an object decreases its motion energy. Increasing the speed of an object increases its motion energy. Increasing the speed of an object does not affect its motion energy. Whether or not its motion energy is affected depends on how much its speed was increased.

Explanation:

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a 700k/g race car slowed down from 30m/s to 15m/s what is the impulse and what was the average force applied by the brakes

posledela

So impulse is a change in momentum.

Mass*(final velocity - initial velocity)

I dont think you will be able to find the average force with the given info because you need to know the time it takes for the car to slow down.

5 0

4 years ago

A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl

Vlada [557]

Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

$\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}$

$R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10$

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

$V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}$

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

$V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}$

Part C

The error in % is given by

$\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%$

4 0

3 years ago

Read 2 more answers

How fast was a plane flying if it traveled 500 km in 30 min?

anyanavicka [17]

Answer:

16.7 km per min

Explanation:

divided 500 and 30

8 0

3 years ago

Read 2 more answers

A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hol

il63 [147K]

Answer:

v = 12.1 m/s

Explanation:

When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.
These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.
In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.
Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.
This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.
At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

$F_{c} = N + m*g (1)$

Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

$F_{c} = m*\frac{v^{2}}{r} (2)$

Since the normal force takes any value as needed to make (1) equal to (2), if the speed diminishes, it will be needed less force to keep the equality valid.
In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.
In this condition, from (1) and (2), we can find the minimum possible value of the speed that still keeps the motorcycle touching the surface, as follows:
$v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)$

6 0

3 years ago

A charged particle is observed traveling in a circular path in a uniform magnetic field. If the particle had been traveling four

8090 [49]

Answer:

The new radius of the trajectory of the particle is four times the previous radius

Explanation:

In order to know what is the radius of the trajectory of the charged particle, if its speed is four times as fast, you take into account the following formula, which describes the radius of a charged particle in a magnetic field:

$r=\frac{mv}{qB}$ (1)

If the speed of the particle is for time as fast, that is, v' = 4v, you obtain, in the equation (1):

$r'=\frac{mv'}{qB}=\frac{m(4v)}{qB}=4\frac{mv}{qB}=4r$

The new radius of the trajectory of the particle is four times the previous radius

8 0

3 years ago