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hammer [34]
2 years ago
5

The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i

s 1.5multiply1011 m. (a) Calculate the magnitude of the gravitational force exerted by the Sun on the Earth. N (b) Calculate the magnitude of the gravitational force exerted by the Earth on the Sun.
Physics
1 answer:
spayn [35]2 years ago
7 0

Answer:

<em>a) 3.56 x 10^22 N</em>

<em>b) 3.56 x 10^22 N</em>

<em></em>

Explanation:

Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = \frac{GMm}{R^2}

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = \frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2} = <em>3.56 x 10^22 N</em>

<em></em>

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to <em>3.56 x 10^22 N</em>

b) The force exerted by the Earth on the Sun = <em>3.56 x 10^22 N</em>

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maxonik [38]

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.

We  also know that with additional streching of 2 cm of teh spring,  the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

k=2*7/(0.02)^2=35000 N/m

7 0
2 years ago
"For a first order instrument with a sensitivity of .4 mV/K and a time" constant of 25 ms, find the instrument’s response as a f
ELEN [110]

Answer:

Explanation:

Given that:

For a first order instrument with a sensitivity of .4 mV/K

constant c  = 25 ms = 25 × 10⁻³ s

The initial temperature T_1 = 273 K

The final temperature T_2 = 473 K

The initial volume = 0.4 mV/K × 273 K = 109.2 V

The final volume =  0.4 mV/K × 473 K =  189.2 V

the instrument’s response as a function of time for a sudden temperature increase can be computed as follows:

Let consider y to be the function of time i.e y(t).

So;

y(t) = 109.2  + (189.2 - 109.2)( 1 - \mathbf{e^{-t/c}})mV

y(t) = (109.2 +  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

Plot the response y(t) as a function of time.

The plot of y(t) as a function of time can be seen in the diagram  attached below.

What are the units for y(t)?

The unit for y(t) is mV.

Find the 90% rise time for y(t90) and the error fraction,

The 90% rise time for y(t90) is as follows:

Initially 90% of 189.2 mV = 0.9 ×  189.2 mV

=  170.28 mV

170.28 mV = (109.2 +  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

170.28 mV - 109.2 mV = 80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

61.08 mV =  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

0.7635  mV = ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

t = 1.44 × 25  × 10⁻³ s

t = 0.036 s

t = 36 ms

The error fraction = \dfrac{189.2-170.28  }{189.2}

The error fraction = 0.1

The error fraction = 10%

8 0
2 years ago
A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r
Firlakuza [10]

Answer:

Speed of the this part is given as

v_3 = \sqrt2 v

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

P_1 + P_2 + P_3 = P_i

here we know that

P_1 = P_2

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

P_1 + P_2 = \sqrt{P^2 + P^2}

P_1 + P_2 = \sqrt2 mv

now from above equation we have

P_3 = -(P_1 + P_2)

mv_3 = -(\sqrt 2 mv)

v_3 = \sqrt2 v

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

6 0
2 years ago
Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly
loris [4]

Answer: 36.86\°

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: d=2.4 m

Ferdinand's initial velocity: V_{o}=5 m/s

Time it takes a jump: t=0.6 s

We need to find the angle \theta at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

V_{x}=\frac{d}{t} (1)

V_{x}=\frac{2.4 m}{0.6 s} (2)

V_{x}=4 m/s (3)

On the other hand, the x-component of the velocity is expressed as:

V_{x}=V_{o}cos\theta (4)

Substituting (3) in (4):

4 m/s=5 m/s cos\theta (5)

Clearing \theta:

\theta=cos^{-1} (\frac{4 m/s}{5 m/s})

\theta=36.86\° This is the angle at which Ferdinand the frog jumps between lily pads

4 0
2 years ago
A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm
Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g

The mass of the meter stick is 126.99115 g

6 0
2 years ago
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