All categories

2 years ago

The gravitational attraction between two masses of 3kg that are separated by a distance of 1cm is

Physics

1 answer:

Luba_88 [7]2 years ago

6 0

Answer:

6.003×10¯⁶ N

Explanation:

We'll begin by converting 1 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

1 cm = 1 cm × 1 m / 100 cm

1 cm = 0.01 m

Finally, we shall determine the gravitational attraction. This can be obtained as follow:

Mass 1 (M₁) = 3 Kg

Mass 2 (M₂) = 3 Kg

Distance apart (r) = 0.01 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 3 × / 0.01²

F = 6.003×10¯¹⁰ / 1×10¯⁴

F = 6.003×10¯⁶ N

Thus the gravitational attraction is 6.003×10¯⁶ N

You might be interested in

A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,

Svetlanka [38]

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

To equate

357.18 m -144.54 m = .735 m d

d = 289.31 m .

4 0

3 years ago

Jesse wants to know how well a particular brand of car wax protects his car from dirt. What is the independent variable ?

miv72 [106K]

The brand of car wax

4 0

2 years ago

A basketball is shot at 14.0 m/s at a 65.0 degree angle. What is the magnitude only (no direction) of the velocity of the ball 2

Alex787 [66]

Answer:

10.4 m/s

Explanation:

I just know it's right!

8 0

3 years ago

A vehicle travelling at an initial velocity of 20km/hr,accelerates at 4m/s².calculate its final velocity after 10 seconds.

Nikitich [7]

acceleration = Velocity changes ÷ time of the velocity changes

4 m/s^2 =

4 × 10^(-3) × 3600 km / h =

4 × 3.6 =

14.4 km / h

Thus :

14.4 = V(2) - V(1) / t(2) - t(1)

14.4 = V(2) - 20 / 10

Multiply both sides by 10

10 × 14.4 = 10 × ( V(2) - 20 ) / 10

144 = V(2) - 20

Add both sides 20

144 + 20 = V(2) - 20 + 20

V(2) = 164 Km/h

Thus the final velocity after 10 seconds is 164 Km/h .

4 0

2 years ago

51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-

Tanya [424]

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance $r= 5.00\ \Omega$

(a). We need to calculate the current

Using rule of loop

$E-IR-Ir=0$

$I=\dfrac{E}{R+r}$

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

$I=\dfrac{3.200}{1.00\times10^{3}+5.00}$

$I=3.184\times10^{-3}\ A$

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

$V=E-Ir$

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

$V=3.200-3.184\times10^{-3}\times5.00$

$V=3.18\ V$

(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

$\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }$

$\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}$

Hence, This is the required solution.

5 0

3 years ago