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3 years ago

13

The force of gravitation between two spherical bodies is Gm1

1 answer:

malfutka [58]3 years ago

3 0

Answer:

r is the separation between the two spherical bodies

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A planet similar to the Earth has a radius 7.4 × 106 m and has an acceleration of gravity of 10 m/s 2 on the planet’s surface. T

Lubov Fominskaja [6]

Answer:

1.5 hr

16.7

Explanation:

Zero apparent weight means there's no normal force.

Sum the forces in the centripetal direction.

∑F = ma

mg = mv²/r

v = √(gr)

v = √(7.4×10⁶ m × 10 m/s²)

v = 8602 m/s

The circumference of the equator is:

C = 2πr

C = 2π (7.4×10⁶ m)

C = 4.65×10⁷ m

So the period is:

T = C / v

T = (4.65×10⁷ m) / (8602 m/s)

T = 5405 s

T = 1.5 hr

The initial speed is:

v = C / T

v = (4.65×10⁷ m) / (25 h × 3600 s/h)

v = 517 m/s

The speed increases by a factor of:

8602 m/s / 517 m/s

16.7

3 0

3 years ago

A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.

sladkih [1.3K]

Explanation:

Given that,

Charge 1, $q_1=6.75\ nC=6.75 \times 10^{-9}\ C$

Charge 2, $q_2=4.46\ nC=4.46\times 10^{-9}\ C$

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

$F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}$

$F=6.84\times 10^{-8}\ N$

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

5 0

4 years ago

The orbital motion of Earth around the Sun leads to an observable parallax effect on the nearest stars. For each star listed, ca

Murrr4er [49]

Answer:

Following are the answer to this question:

Explanation:

Formula:

$D(PC) =\frac{1}{parallax}\\\\D(av)=D(PC) \times 20.626\ J$

Calculating point A:

when the value is $0.38$

$\to 0.38 \toD(PC)= \frac{1}{0.38}\\\\$

$=2.632$

$\to D(a.v) = \frac{1}{0.38} \times 206265\\$

$=542,802.6$

Calculating point B:

when the value is $0.75$

$\to D(PC)=\frac{1}{0.75}$

$=1.33$

$\to D(a.v) = \frac{1}{0.75} \times 206265\\$

$=275,020$

Calculating point C:

when the value is $0.28$

$\to D(PC)=\frac{1}{0.28}$

$=3.571$

$\to D(a.v) = \frac{1}{0.28} \times 206265\\$

$=736660.7$

Calculating point D:

when the value is $0.42$

$\to D(PC)=\frac{1}{0.42}$

$=2.38$

$\to D(a.v) = \frac{1}{0.42} \times 206265\\$

$=490910.7$

Calculating point E:

when the value is $0.31$

$\to D(PC)=\frac{1}{0.31}$

$=3.226$

$\to D(a.v) = \frac{1}{0.31} \times 206265\\$

$=665370.97$

6 0

3 years ago

A surface completely surrounds a 3.3 × 10-6 C charge. Find the electric flux through this surface when the surface is (a) a sphe

KengaRu [80]

Answer:

Electric flux in a) , b) and c) is same which is 0.373 × 10 ⁶ N m²/C

Explanation:

given,

surface charge (q) = 3.3 × 10⁻⁶ C

to calculate electric flux = ?

a) radius = 0.76 m

area of sphere = 4 π r²

electric flux = $\dfrac{q}{\varepsilon}$

$\varepsilon = 8.85 \times 10^{-12} C^2/Nm^2$

electric flux = $\dfrac{3.3 \times 10^{-6}}{8.85 \times 10^{-12} }$

flux = 0.373 × 10 ⁶ N m²/C

electric flux in the other two cases will also be same as electric flux is independent of area

so, Electric flux in a) , b) and c) is same which is 0.373 × 10 ⁶ N m²/C

5 0

4 years ago

What is the objects acceleration

alekssr [168]

Answer:

Acceleration is a vector quantity which is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.

5 0

2 years ago

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