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3 years ago

What other factor affected the gravitational force between the two objects in this simulation?

Physics

1 answer:

neonofarm [45]3 years ago

7 0

Answer:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases

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A piece of wire length 30cm and cross sectional area of 0.5mm^2 has a resistance of 5ohms at 20°c. It is then heated to a temper

Grace [21]

Answer:

8.333*10^-6 ohms

Explanation:

Resistivity of a material is expressed as;

p = RA/l

R is the resistance of the material

A is the cross sectional area

l is the length of the material

Given

R = 5 ohms

A = 0.5mm^2

A = 5 * 10^-7m^2

l = 30cm = 0.3m

Substitute into the formula;

p = (5 * 5 * 10^-7m^2)/0.3

p = 25 * 10^-7/0.3

p = 0.0000025/0.3

p = 8.333*10^-6

Hence its resistivity at 20 degrees Celsius is 8.333*10^-6 ohms

7 0

3 years ago

One heater uses 310 W of power when connected by itself to a battery. Another heater uses 180 W of power when connected by itsel

igor_vitrenko [27]

Answer:

55.8W

Explanation:

P= V^2/R

R= V^2/P

For series connection

Req= R1+ R2= V^2/310 + V^2/180

R=V^2/P= V^2/310 + V^2/180

But V^2 will cancel out

P= 1/(1/310 + 1/180)

P= 55.8W

8 0

2 years ago

Read 2 more answers

Two tiny, spherical water drops, with identical charges of −8.00 ✕ 10^(−17) C, have a center-to-center separation of 2.00 cm.

Damm [24]

Answer:

F=1.4384×10⁻¹⁹N

Explanation:

Given Data

Charge q= -8.00×10⁻¹⁷C

Distance r=2.00 cm=0.02 m

To find

Electrostatic force

Solution

The electrostatic force between between them can be calculated from Coulombs law as

$F=\frac{kq^{2} }{r^{2} }$

Substitute the given values we get

$F=\frac{(8.99*10^{9} )*(-8.00*10^{-17} )^{2} }{(0.02)^{2} }\\ F=1.4384*10^{-19} N$

7 0

3 years ago

A solid block of mass m2 = 1.14 kg, at rest on a horizontal frictionless surface, is connected to a relaxed spring (with spring

borishaifa [10]

Answer:

v = 1 m/s

Explanation:

from the principle of conservation of momentum, we have following relation

initial momentum = final momentum

$m_{1}v_{1}+m_{2}v_{2} = (m_{1}+m_{2})v^{2}$

where

m1 = 1.14 kg

v1 = 2.0 m/s

m2 = 1.14 kg

v2 = 0 m/s

putting all value in the above equation

$1.14 *2.0+ 0 =(1.14+1.14)v^{2}$

$v =\frac{1.14*2.0}{1.14+1.14}$

v = 1 m/s

4 0

3 years ago

Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

ozzi

Answer:

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

T Cosθ = mg .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

$tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}$

$tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}$

$tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}$

$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

7 0

2 years ago