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lukranit [14]
2 years ago
15

Angelina jumps off a stool. As she is falling, the Earth’s gravitational force on her is larger in magnitude than the gravitatio

nal force she is exerting on the earth. 1. False 2. This may happen in many cases, but not in every case. 3. True 4. None of these 5. This may happen in a few cases, but it is very rare.
Physics
2 answers:
valentinak56 [21]2 years ago
8 0
The answer is True. The amount force exerted by any object is directly proportional to its mass. This means that our planet is exerting more gravitational force to Angelina, and Angelina is also exerting a gravitational force on our planet directly proportional to her mass. Angelina is actually falling towards the center of the earth,and also our planet is also moving towards Angelina, but it seems negligible with respect to Angelina.Our Sun is so massive that it held our planet in its orbit because of its gravitational force.
STatiana [176]2 years ago
5 0

Answer:

1. False

Explanation:

As per law of gravitational attraction force we know that

F = \frac{Gm_1m_2}{r^2}

where we know that

m_1,m_2 = masses of two point mass

r = distance between two masses

so here we know that as the force does not depend on individual mass but it will depend on the product of two masses

so here the gravitational force is same in magnitude on both the bodies

so when Angelina jumps off the stool the gravitational force on Earth and on Angelina must be of same magnitude

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Given the distance r = 2/1000 m, the force between them F = 0.0104 N, the mass of the two object can be calculated using formula:

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And where G = is the gravitational constant (6.67E-11 m3 s-2 kg-1)

The mass of the two objects are 24.96 kg

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Rhea kicks a soccer ball at 13 km/h to Sean. After kicking the ball, the speed of the soccer ball from Rhea’s reference frame is
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Explanation: Your welcome :)

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2 years ago
the grid in a triode is kept negatively charged to prevent… a. the variations in voltage from getting too large. b. electrons be
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D:the electrons from being attracted to the grid instead of the anode
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Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p
worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

  • U1/U2=6
  • U1/U2=6

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

U=\frac{-k*qQ}{r}

Therefore

The Equations of U1 and U2 is

For U1

U1=\frac{-k*q_1Q}{r}

For U2

U2=\frac{-k*q_1Q}{3*2r}

Since

U is a function of q and  q2=q1/3

Therefore

U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

7 0
1 year ago
A particle with a charge of -1.24 x 10"° C is moving with instantaneous velocity (4.19 X 104 m/s)î + (-3.85 X 104 m/s)j. What is
astra-53 [7]

Answer:

(a) F= 6.68*10¹¹⁴ N (-k)

(b) F =( 6.68*10¹¹⁴ i  + 7.27*10¹¹⁴ j  ) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q* v X B Formula (1 )

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q= -1.24 * 10¹¹⁰ C

v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j

B  =(1.40 T)i  

B  =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

            = - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)

1T= 1 N/ C*m/s

We apply the formula (1)

  F= 1.24 * 10¹¹⁰ C*  5.39* 10⁴ m/s* N/ C*m/s (-k)

   F= 6.68*10¹¹⁴ N (-k)

a)  vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =

             =( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C*  (  5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s

F =( 6.68*10¹¹⁴  i  + 7.27*10¹¹⁴  j  ) N

8 0
2 years ago
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