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3 years ago

As the distance between 2 objects increases, what happens to gravitational force?

Physics

1 answer:

ExtremeBDS [4]3 years ago

3 0

Answer:

the force decreases

Explanation:

Gravitational force is depending on it's mass and the distance between two object

The attachment shows us to understand relationship of it's

F ∝ 1/ (r)2

there for gravitational force is inversely proportional to it's distance . when distance between two object is increased, Then force will be decreased

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a motorcar is moving with a velocity of 108 km / h and it takes 4s to stop after the brakes are applied calculate the force exer

k0ka [10]

Answer: - 7500N

Explanation:

Given the following :

Initial Velocity of car = 108km/hr

Time taken to stop after applying brakes = 4s

Mass of passengers in car = 1000kg

Force exerted by the brakes on the car =?

After 4s, then final Velocity (V) = 0

Initial Velocity (u) of the car = 108km/hr

108km/hr = (108 × 1000)m ÷ (3600)s = 30m/s

Force exerted = mass(m) × acceleration(a)

Acceleration of car = Change in Velocity with time

a = (v - u) / t

a = (0 - 30) / 4

a = - 30/ 4

a = - 7.5m/s^2

Therefore,

Force exerted = mass(m) × acceleration(a)

Force exerted = 1000kg × (-7.5)m/s^2

Force exerted = - 7500N

5 0

3 years ago

Read 2 more answers

The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the

FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

$B=\frac{2\pi*a }{u*I}$

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

$I=\frac{2\pi*a*B }{u}$ Formula(2)

$B=0.2*10^{-5} T = 0.2*10^{-5} \frac{weber}{m^{2} }$

a =8cm=0.08m

$u=4*\pi *10^{-7} \frac{Weber}{A*m}$

We replace the known information in the formula (2)

$I=\frac{2\pi*0.08*0.2*10^{-5} }{4\pi *10x^{-7} }$

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0

3 years ago

the maximum intensity levels of a trumpet, trombone, and a bass drum, each at a distance of 3m are 94 dB, 107dB, and 113dB respe

Gwar [14]

Answer:

β = 114 db

Explanation:

The intensity of sound in decibles is

β = 10 log $\frac{I}{I_{o}}$

in most cases Io is the hearing threshold 1 10-12 W / cm²

let's calculate the intensity of each instrument

I / I₀ = 10 (β / 10)

I = I₀ 10 (β / 10)

trumpet

I1 = 1 10⁻¹² 10 (94/10)

I1 = 2.51 10⁻³ / cm²

Thrombus

I2 = 1 10⁻¹² 10 (107/10)

I2 = 5.01 10-2 W / cm²

low

I3 =1 1-12 (113/10) W/cm²

I3 = 1,995 10-1 W / cm²

when we place the three instruments together their sounds reinforce

I_total = I₁ + I₂ + I₃

I_ttoal = 2.51 10-3 + 5.01 10-2 + 1.995 10-1

I_total = 0.00251 + 0.0501 + 0.1995

I_total = 0.25211 W / cm²

let's bring this amount to the SI system

β = 10 log (0.25211 / 1 10⁻¹²)

β = 114 db

7 0

3 years ago

The efficiency of a carnot cycle is 1/6. If on reducing the temperature of the sink 75 degree Celsius, the efficiency becomes 1/

svp [43]

Answer:

375 and 450

Explanation:

The computation of the initial and the final temperature is shown below:

In condition 1:

The efficiency of a Carnot cycle is $\frac{1}{6}$

So, the equation is

$\frac{1}{6} = 1 - \frac{T_2}{T_1}$

For condition 2:

Now if the temperature is reduced by 75 degrees So, the efficiency is $\frac{1}{3}$

Therefore the next equation is

$\frac{1}{3} = 1 - \frac{T_2 - 75}{T_1}$

Now solve both the equations

solve equations (1) and (2)

$2(1 - T_2/T_1) = 1 - (T_2 - 75)/T_1\\\\2 - 1 = 2T_2/T_1 - (T_2 - 75)/T_1\\\\ = (T_2 + 75)/T_1T_1 = T_2 + 75\\\Now\ we\ will\ Put\ the\ values\ into\ equation (1)\\\\1/6 = 1 - T_2/(T_2 + 75)\\\\1/6 = (75)/(T_2 + 75)$

T_2 + 450 = 75

T_2 = 375

Now put the T_2 value in any of the above equation

i.e

T_1 = T_2 + 75

T_1 = 375 + 75

= 450

7 0

4 years ago

An Olympic-class sprinter starts a race with an acceleration of 5.10 m/s2. What is her speed 2.40 s later?

ivolga24 [154]

Answer:

12.24 m/s

Explanation:

Speed: This can be defined as the rate of change of distance with time. The S.I unit of speed is m/s.

Using the formula,

a = v/t................ Equation 1

Where a = acceleration of the sprinter, v = speed of the sprinter, t = time.

making v the subject of the equation,

v = at ................. Equation 2

Given: a = 5.1 m/s², t = 2.4 s.

Substitute into equation 2

v = 5.1(2.4)

v = 12.24 m/s.

Hence, the speed of the sprinter = 12.24 m/s

3 0

3 years ago