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Vinil7 [7]
2 years ago
13

What is the magnitude of the gravitational force acting on the sun due to the earth?The earth does not exert any gravitational f

orce on the sun._______________The earth exerts some force on the sun, but less than 3.53×1022N because the earth, which is exerting the force, is so much less massive than the sun._______________The earth exerts 3.53×1022N of force on the sun, exactly the same amount of force the sun exerts on the earth found in Part A._______________The earth exerts more than 3.53×1022N of force on the sun because the sun, which is experiencing the force, is so much more massive than the earth._______________
Physics
1 answer:
Sholpan [36]2 years ago
4 0

I haven't worked on Part-A, and I don't happen to know the magnitude of the gravitational force that the Sun exerts on the Earth.  

But whatever it is, it's exactly, precisely, identical, the same, and equal to the magnitude of the gravitational force that the Earth exerts on the Sun.

I think that's the THIRD choice here, but I'm not sure of that either.

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Solve 3x+1<br><img src="https://tex.z-dn.net/?f=%20%5Cleqslant%201" id="TexFormula1" title=" \leqslant 1" alt=" \leqslant 1" ali
Rzqust [24]

3x + 1 ≤ 1

Subtract 1 from each side:  3x ≤ 0

Divide each side by  3 :     <em>x ≤ 0</em>

8 0
2 years ago
Two boxes are connected to each other by a string as shown in the figure. The 10-n box slides without friction on the horizontal
FromTheMoon [43]

Answer:

T=7.4 N hence T<30 N

Explanation:

The figure is likely to be similar to the one attached. Writing the equation for forces we have

F-T=Fa/g where F is the force, T is tension, a is acceleration and g is acceleration due to gravity. Substituting the figures we have the first equation as

30 N - T = (30/9.81)a

Also, we know that T=F*a/g and substituting  10N for F we obtain the second equation as

T = (10/9.81)a

Adding the first and second equations we obtain

30 = 4.077471967

a Hence

a=\frac {30}{4.077471967}=7.3575 m/s^{2}

and T=a hence

T is approximately 7.4 N

5 0
3 years ago
Read 2 more answers
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
xz_007 [3.2K]

Answer:

a) \Delta{t} = 5.39s

b) the motorcycle travels 155 m

Explanation:

Let t_2-t_1 = \Delta{t}, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}

where:

v_{m2} is the speed of the motorcycle at time 2

v_{c} is the velocity of the car (constant)

v_{0} is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]

v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933

3 0
3 years ago
You are riding a bicycle up a gentle hill. It is fairly easy to increase your potential
Dahasolnce [82]
True or false: while riding a bicycle up a gentle hill, it fairly easy to increase your potential energy, but to increase your kinetic energy would ...
3 0
1 year ago
A very long train is rolling at 4 m/s along a straight track. An observer is standing on the ground very dangerously close to th
hichkok12 [17]

Answer:

A. \vec{r}=(6\frac{m}{s})t\ \ \hat{i}

B.  t = 50 s

Explanation:

A. The vectorial equation of the person who is getting closer to the other person is:

\vec{r}=\vec{v}t

r: position vector

v: speed vector = 6m/s i  (if you consider the motion as a horizontal motion)

Then, you replace and obtain:

\vec{r}=(6\frac{m}{s})t\ \ \hat{i}

B. The time is:

t=\frac{d}{v}

d: distance to the observer = 300m

v: speed of the person on the car = 6.00 m/s

t=\frac{300m}{6m/s}=50s

4 0
3 years ago
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