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4 years ago

A positively charged object will attract an object that has

Physics

1 answer:

ser-zykov [4K]4 years ago

5 0

I’d say a negative charge cause opposites attract but I’m not sure

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A baseball player slides into third base with an initial speed of 4.0 m/s . if the coefficient of kinetic friction between the p

satela [25.4K]

The work done by the friction force to stop the player is equal to his loss of kinetic energy:
$W= \Delta K$

The work done by the friction force is the magnitude of the force $\mu m g$ times the distance covered by the player, d:
$W= \mu mg d$

The loss in kinetic energy is simply equal to the initial kinetic energy of the player, since the final kinetic energy is zero (the player comes to rest):
$\Delta K=K_i = \frac{1}{2}mv^2$

Substituting into the first equation, we get:
$\mu m g d= \frac{1}{2}mv^2$
from which we find d, the distance covered by the player:
$d= \frac{v^2}{2 \mu g}= \frac{(4.0 m/s)^2}{2 \cdot 0.4 \cdot 9.81 m/s^2}=2.04 m$

4 0

3 years ago

What are the key differences between stored and conservation of energy?

Irina18 [472]

Stored energy=energy that is stored and may not be used, conservation of energy= saved energy that you can use for other things.

7 0

3 years ago

An initially motionless test car is accelerated uniformly to 115 km/h in 8.83 seconds before striking a simulated deer. The car

goblinko [34]

We know the equation of motion v = u+ at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

In this case Final velocity before collision = 115 km/hr = 115*5/18 = 31.94 m/s

Time taken by car to reach this velocity = 8.83 seconds

Initial velocity = 0 m/s

v = u +at

31.94 = 0 + a*8.83

a = 3.62 $m/s^2$

So acceleration of car just before collision = 3.62 $m/s^2$

3 0

4 years ago

8_murik_8 [283]

Answer: 2.61 s

Explanation:

We are given the following data:

$s=5 ft$ is the initial height of the object

$v=40 ft/s$ is the initial height of the object

In addition, the motion of the object is given by:

$h=-16t^{2}+ vt +s$ (1)

Where:

$h=0 ft$ is the final height of the object

$t$ is the time the object is in the air before hitting the ground

Rewritting (1) with the given data:

$0=-16t^{2}+ 40t +5$ (2)

Solving the quadratic equation with the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ , where $a=-16$ , $b=40$ , $c=5$ and choosing the positive value of time:

$t=\frac{-40\pm\sqrt{(-40)^{2}-4(-16)(5)}}{2(-16)}$ (3)

$t=2.61 s$ (4) This is the time

5 0

4 years ago

A heater with an effective resistance of 4.8 ohms is connected to 240.0 volts if it operated approximately 25% of the time and e

agasfer [191]

-- Power used when the heater is running =

 V²/R = (240)²/4.8= 12,000 watts. (12 KW)

(A gigantic heater ... suitable for maybe a theater, not for a home.)

-- Energy consumed in one day:

 (12 KW) x (24 hours) x (25%) = 72 KwH per day

-- Energy consumed in 30 days: (72 KwH/day) x (30 days) =

 2,160 KwH in 30 days

-- Cost = ($0.1 / KwH) x (2,160 KwH) = $216.00 in 30 days

6 0

3 years ago