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2 years ago

13

________________________ energy is the energy that an object gets from its position and motion. Question 2 options: mechanical n

uclear electrical none of the above

1 answer:

shutvik [7]2 years ago

3 0

I think it’s kinetic

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Find the escape velocity ve for an object of mass m that is initially at a distance r from the center of a planet of mass m. ass

skelet666 [1.2K]

We are given
m = mass of the object
r = distance from the center of the planet
r <span>≥ rplanet

We are asked for the
ve = escape velocity

The escape velocity
ve = </span>√ (2Gm/r)
<span>
</span>

4 0

3 years ago

Vesna Vulovic survived the longest fall on record without a parachute when her plane exploded and she fell 5 miles, 733 yards. W

Valentin [98]

Answer:

8717 meters.

Explanation:

We need to know the conversion factors. We know that:

1 mile = 1609.34 meters

1 yard = 0.9144 meters

This means that:

$\frac{1609.34 meters}{1 mile}=1$

$\frac{0.9144 meters}{1 yard}=1$

It is convenient to leave the units we want at the end in the numerator so the ones in the denominator cancel out with the ones we want to remove, as will be seen in the next step.

We will convert first the miles, then the yards, and add them up.

$5miles=5miles\frac{1609.34 meters}{1 mile}=8046.7meters$

$733yards=733yards\frac{0.9144 meters}{1 yard}=670.2552meters$

So total distance is the sum of these, 8717 meters.

8 0

3 years ago

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s,

Art [367]

Answer:

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

Explanation:

given,

length of ladder = 10 ft

let x be the distance of the bottom and y be the distance of the top of ladder.

x² + y² = 100

differentiating with respect to time we get

$2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0$ ..............(1)

when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s

from equation (1)

now,

$16\times 1.4 + 12\dfrac{dy}{dt} = 0$

$\dfrac{dy}{dt} = -\dfrac{5.6}{3}$

let the angle between the ladders be θ

$tan\theta = \dfrac{y}{x}$

y = xtan θ

$\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt}$

$-\dfrac{5.6}{3} =1.4\times \dfrac{6}{8} + 8 (1+\dfrac{9}{16})\dfrac{d\theta}{dt}$

$\dfrac{25}{2} \dfrac{d\theta}{dt} =\dfrac{-17.5}{6}$

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

6 0

2 years ago

A motorcycle of mass 250 kg drives around a circle with a centripetal acceleration of 3.4 m/s2. What is the centripetal force ac

Vlad [161]

Hope this helps and have a nice day!

8 0

2 years ago

Read 2 more answers

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

2 years ago