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3 years ago

13

the gravitational pull of the moon is much less than the gravitational pull of earth, which two statements are true for an objec

t with a mass of 20 kilograms that weighs 44 pounds on earth

1 answer:

alexgriva [62]3 years ago

6 0

Answer:

b

Explanation:

the earths mass is more than the moon .

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The volume V of a right circular cylinder of radius r and height h is V=πr2h. (a) How is dVdt related to drdt if h is constant a

Nikolay [14]

Answer:

(a) $\frac{dV}{dt}= 2\pi r h \frac{dr}{dt}$

(b) $\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

(c) $\frac{dV}{dt} = \pi r^2\frac{dh}{dt}+2\pi r h\frac{dr}{dt}$

Explanation:

Differentiating Rules:

$\frac{dx^n}{dx}= nx^{n-1}$
$\frac{dx}{dx}=1$
$\frac{d}{dx}(mn)= m\frac{dn}{dx}+n\frac{dm}{dx}$ [ m and n are the function of x]
$\frac{d}{dx}(cn)=c \frac{dn}{dx}$ [ here c is constant and n is function of x]

Given that,

$V= \pi r^2h$

(a)

$V= \pi r^2h$

Differentiating with respect to t

$\frac{dV}{dt}= \frac{d}{dt}(\pi r^2h)$

$\Rightarrow \frac{dV}{dt}= \pi h \frac{d}{dt}(r^2)$ [ here $\pi h$ is constant]

$\Rightarrow \frac{dV}{dt}= \pi h 2r \frac{dr}{dt}$

$\Rightarrow \frac{dV}{dt}= 2\pi r h \frac{dr}{dt}$

(b)

$V= \pi r^2h$

Differentiating with respect to t

$\frac{dV}{dt}= \frac{d}{dt}(\pi r^2h)$

$\Rightarrow \frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

(c)

$V= \pi r^2h$

Differentiating with respect to t

$\frac{dV}{dt}= \frac{d}{dt}(\pi r^2h)$

$\Rightarrow \frac{dV}{dt} = \pi r^2\frac{dh}{dt}+\pi h\frac{d}{dt}(r^2)$

$\Rightarrow \frac{dV}{dt} = \pi r^2\frac{dh}{dt}+2\pi r h\frac{dr}{dt}$

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4 years ago

A pendulum is observed to complete 23 full cycles in 58 seconds. Use the definition of frequency to find the frequency.

Alex17521 [72]

Answer: f = 0.397 Hz

Explanation:

f = cycles per second

f = 23 / 58 = 0.39655...

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3 years ago

Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and init

yulyashka [42]

To solve this problem we will apply the conventions related to the cinematic movement theorem, for which the kinematic equations of linear motion are equally detached. We will use the speed and position equations to determine the general formula according to the given values.

To velocity function we have

$v(t) = v(0)+\int^t_0 a(x) dx$

Our values are,

$a(x) = -9.8$

$v(0) = 20$

$s(0) = 0$

Replacing at this equation and solving we have that the equation for the velocity would be,

$v(t) = 20+\int^t_0 -9.8 dx$

$v(t) = 20-\int^t_0 9.8 dx$

$v(t) = 20-9.8x|^t_0$

$v(t) = 20-9.8(t-0)$

$v(t) = 20-9.8t$

Therefore the velocity function is $v(t) = 20-9.8t$

At the same time for the position function:

$s(t) = s(0) +\int^t_0 v(x) dx$

Replacing we have that

$s(t) = s(0) +\int^t_0 20-9.8x dx$

$s(t) = 0 +20x-9.8\frac{x^2}{2}|^t_0$

$s(t) = (20t-9.8\frac{t^2}{2})-(20\cdot 0 -9.8 \frac{0^2}{2})$

$s(t) = 20t-9.8\frac{t^2}{2}$

Therefore the position function is

$s(t) = 20t-9.8\frac{t^2}{2}$

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3 years ago

A raindrop has a mass of 7.7 × 10-7 kg and is falling near the surface of the earth. Calculate the magnitude of the gravitationa

goldfiish [28.3K]

Given that the mass of the raindrop is

$m=7.7\times10^{-7}\text{ kg}$

The acceleration due to gravity is g = 9.81 m/s^2

We have to find

(a) Magnitude of the gravitational force exerted on the raindrop by earth

(b)Magnitude of the gravitational force exerted on the earth by the raindrop

(a) The formula to calculate the magnitude of the gravitational force exerted on the raindrop by the earth is

$F=mg$

Substituting the values, the gravitational force exerted on the raindrop by the earth is

$\begin{gathered} F=7.7\times10^{-7}\times9.81 \\ =7.55\times10^{-6}\text{ N} \end{gathered}$

(b) According to Newton's third law,

If F' is the gravitational force exerted on the earth by the raindrop, then

$F=-F^{\prime}$

Here, the negative sign indicates that both forces act in opposite direction.

The gravitational force exerted on the earth by the raindrop is

$F^{\prime}\text{ = -7.55}\times10^{-6}\text{ N}$

And the magnitude of the gravitational force exerted on the earth by the raindrop is

$7.55\times10^{-6}\text{ N}$

Thus, the magnitude of both forces is equal.

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1 year ago

First Law of Thermodynamics - Sign Convention The first law of thermodynamics applies the conservation of energy principle to sy

insens350 [35]

The question is incomplete. The complete question is :

First Law of Thermodynamics - Sign Convention The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is AU = Q-W. Here AU is the change in internal energy U of the system. Q is the net heat transferred into the system that is, Q is the sum of all heat transfer into (positive) and out of (negative) the system. W is the net work done by the system—that is, W is the sum of all work done by (positive) and on (negative) the system. We use the following sign conventions: if Q is positive, then there is a net heat transfer into the system; if W is positive, then there is net work done by the system. So positive Q adds energy to the system and positive W takes energy from the system. Thus AU = Q-W. Note also that if more heat transfer into the system occurs than work done, the difference is stored as internal energy. The first law of thermodynamics AU = 9 - W Ur-U, Heat Work System AU--W system w Qin: talu Qout:

The first law of thermodynamics AU = Q - W U-U Heat Work System AUQ-W Qin: ta Qout: - Wout: + WK w Win: - - volume expands t volume decreases o All answers can be positive or negative. (a) Suppose there is heat transfer of 42 ) into a system, while the system does 6 ) of work. Later, there is heat transfer of 22 J out of the system while 6 ) of work is done on the system. What is the net heat transfer? 20 Correct (100.0%) Submit What is the total work? Enter a number Submit (5 attempts remaining) What is the net change in internal energy of the system? Enter a number

What is the net change in internal energy of the system? Enter a number Submit (5 attempts remaining) (b) What is the change in internal energy of a system when a total of 140 J of heat transfer occurs out of (from) the system and 165 ) of work is done on the system? Enter a number Submit (5 attempts remaining) (c) An athlete doing push-ups performs 645 kJ of work and loses 440 kJ of heat. What is the change in the internal energy (in kJ) of the athlete? Enter a number Submit (5 attempts remaining) kJ (d) An athlete doing push-ups performs 690 kJ of work and loses 450 kJ of heat. Then he takes in 830 kJ of energy from eating food, What is the total change in the internal energy (in kJ) of the athlete? Enter a number kJ.

Solution :

a). Given :

$Q_1 = 42 \ J$ , $Q_2 = -22 \ J , \ W_1 = 6 \ J, \ W_2 = -6 \ J$

Net heat transfer

$Q= Q_1+Q_2$

= 42 + (-22)

= 20 J

Total work

$W= W_1+W_2$

= 6 + (-6)

= 0 J

∵ ΔU = Q - W

= 20 - 0

= 20 J

This is the net change in the internal energy of the system.

b). ΔU = Q + W

= (-140) + (-165)

= -305 J

c). ΔU = Q + W

= (-440) + (645)

= 205 J

d). ΔU = Q + W

= (-450) + (690)

= 240 J

3 0

3 years ago