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Nataly_w [17]
3 years ago
8

Big Ben, a large artifact in England, has a mass of 1x10^8 kilograms and the Empire State Building 1x10^9 kilograms. The distanc

e between them is about 5,000,000 meters. What is the force exerted by Big Ben on the Empire State Building?
Physics
1 answer:
TiliK225 [7]3 years ago
3 0

Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

The mass of Big Ben, M₁ = 1 × 10⁸ kg

The mass of the Empire State Building, M₂ = 1 × 10⁹ kg

The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

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Kipish [7]

Answer:

Velocity

Explanation:

"The principle is that the slope of the line on a position-time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s."

^^This explanation is from physicsclassroom.com

3 0
3 years ago
A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene
Yuri [45]

Explanation:

The given data is as follows.

            U_{a} = 5.4 \times 10^{-8} J

        W_{/text{a to b}} = -1.9 \times 10^{-8} J

        Electric potential energy (U_{b}) = ?

Formula to calculate electric potential energy is as follows.

            U_{b} = U_{a} - W_{/text{a to b}}

                        = 5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)

                        = 7.3 \times 10^{-8} J

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is 7.3 \times 10^{-8} J.

6 0
3 years ago
A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo
Alla [95]

Answer:

v_{ox}= 19.6\ m/s

Explanation:

Data provided in the question:

Height above the ground, H= 5.0m

Range of the ball, R= 20 m

Initial horizontal velocity = v_{ox}

Initial vertical velocity= v_{oy}  (Since ball was thrown horizontally only)

Acceleration acting horizontally, a_x = 0 m/s²  [ Since no acceleration acts horizontally) ]

Vertical Acceleration, a_y = 9.8 m/s² (Since only gravity acts on it)

Let 't' be the time taken to reach ground

Therefore, using equations of motion, we have

H= v_{oy}t+\frac{1}{2}a_yt^2

5= (0)t+\frac{1}{2}(9.8)t^2

t= \frac{10}{9.8}=1.02 s

Then using Equations of motion for horizontal motion,

R= v_{ox}t+\frac{1}{2}a_xt^2

20= v_{ox}(1.02)+\frac{1}{2}(0)(1.02)^2

v_{ox}= 19.6\ m/s

4 0
3 years ago
20. Using the picture, how many neutrons does lithium have?
Igoryamba

Answer:

No. of Neutrons = 3

Explanation:

The atomic number of Lithium is given as 3 in the symbol while the mass number is given as 5.941 which is approximately equal to 6.

Mass Number = No. of Protons + No. of Neutrons = 6

Atomic Number = Number of Electrons = No. of Protons = 3

Therefore,

Mass Number - Atomic Number = (No. of Protons + No. of Neutrons) - No. of Protons

Mass Number - Atomic Number = No. of Neutrons

No. of Neutrons = 6 - 3

<u>No. of Neutrons = 3 </u>

8 0
3 years ago
A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with
Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)

Charge density of the shell is,

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}

Here, R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}

B)

The electric field is E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

C)

For R <r <2R According to gauss law

E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)

substitute \rho=\frac{-3Q}{28\piR^3}

E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}

D)

The net charge enclosed for each r in this range is positive and the electric field is outward

E)

For r>2R

Charge enclosed is zero, so electric field is zero

8 0
3 years ago
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