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3 years ago

8

How much energy Δ???? would be required to move the Moon from its present orbit around Earth to a location that is twice as far

away? Assume the Moon’s orbit around Earth is nearly circular and has a radius of 3.84×108 m. Δ

1 answer:

Zinaida [17]3 years ago

7 0

Answer:

$W = 1.9 \times 10^{28} J$

Explanation:

As we know that total energy of satellite is given as

$E = -\frac{GM_1M_2}{2r}$

now we know that work done to change the orbit is equal to change in the total energy of Earth - Moon system

so we know that

$W = E_f - E_i$

$W = -(\frac{GM_1M_2}{2r_2}) + (\frac{GM_1M_2}{2r_1})$

$W = \frac{GM_1M_2}{2}(\frac{1}{r_1} - \frac{1}{r_2})$

$W = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})(5.98 \times 10^{24})}{2}(\frac{1}{3.84 \times 10^8} - \frac{1}{7.68 \times 10^8})$

$W = 1.9 \times 10^{28} J$

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How do you calculate the braking distance

Anettt [7]

The braking distance is given by $s=\frac{-u^2}{2a}$

Explanation:

When the driver of a car hits the pedal of the brakes, the car starts decelerating until it stops. Assuming the deceleration is constant, then the motion is a uniformly accelerated motion, so we can use the following suvat equation:

$v^2-u^2=2as$

where

u is the initial speed of the car

v is the final speed of the car, which is zero because the car comes to rest:

v = 0

a is the acceleration of the car

s is the distance travelled by the car during the deceleration, so it is the braking distance

Therefore, re-arranging the equation for s, we find an expression for the braking distance:

$s=\frac{-u^2}{2a}$

Note that the sign of $a$ is negative since the car is decelerating, therefore the final sign of $s$ is positive.

Learn more about accelerated motion:

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4 years ago

If |v|= 11, |w|= 23, |v-w|= 30, find |v+w|

GenaCL600 [577]

Answer:

$|v+w|=20$

Explanation:

We are given that

|v|=11

|w|=23

|v-w|=30

We have to find the value of |v+w|

|a-b|^2=(a+b)\cdot (a+b)=a^2+b^2-2|a||b|cos\theta

Using the formula

$(30)^2=(11)^2+(23)^2-2(11)(23)cos\theta$

$900=121+529-506cos\theta$

$900-121-529=-506cos\theta$

$250=-506cos\theta$

$cos\theta=-\frac{250}{506}$

$|a+b|^2=|a|^2+|b|^2+2a\cdot bcos\theta$

Using the formula

$|v+w|^2=(11)^2+(23)^2+2(11)(23)\times (-\frac{250}{506})$

$|v+w|^2=400$

$|v+w|^2=(20)^2$

$|v+w|=20$

7 0

3 years ago

Why do our bodies age?

marysya [2.9K]

When our body reduces Collagen with ageing then, our bodies get aged

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4 years ago

Air having a pressure of 40 psig and a volume of 8 cu ft expands isothermally to a pressure of 10 psig. Find the external work p

DerKrebs [107]

Answer:

357.6 lb-ft

Explanation:

V = Volume = 8 ft³

dP = Change in pressure = (40-10) = 30 psig

Work done is given by

$W=VdP\\\Rightarrow W=8\times (40-10)\\\Rightarrow W=240\ psig.ft^3$

$30\ psig=44.7\ psi\\\Rightarrow 1\ psi=\dfrac{30}{44.7}$

So, converting to ft-lb

$\dfrac{240}{\dfrac{30}{44.7}}=357.6\ lb-ft$

The external work performed during the expansion is 357.6 lb-ft

7 0

4 years ago

If in the experiment, m1 is 38 g, m2 is 63 g, and m3 is 58 g, "and m3 remains static, what is the tension in the string connecti

melomori [17]

Let's call a the acceleration of the system. The problem says that the block m3 is static, so the acceleration is zero: a=0.
Calling $T_1$ the tension of the string between m1 and m3, and $T_2$ the tension of the string between m2 and m3, the problem can be solved by writing the following system of equations:
$m_1 g-T_1=m1_a$
$T_1-T_2=m_3 a$
$T_2-m_2g=m_2a$
However, we know that a=0 and the problem asks only for $T_1$ , so we just need to solve the first equation:
$m_1 g -T_1 =0$
and so
$T_1=m_1g=0.038~kg \cdot 9.81~m/s^2 = 0.37~N$

7 0

3 years ago