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aksik [14]
3 years ago
10

if you have a mass of 55 kg and you are standing 3 meters away from your car, which has a mass of 1234 kg, how strong is the for

ce of gravity between you and the car?
Physics
1 answer:
bagirrra123 [75]3 years ago
5 0

Gravitational force between two masses is given by formula

F = \frac{Gm_1m_2}{r^2}

here we know that

m_1 = 55 kg

m_2 = 1234 kg

r = 3 m

G = 6.67 \times 10^{-11} Nm^2/kg^2

now from the above equation we will have

F = \frac{(6.67 \times 10^{-11})(55)(1234)}{3^2}

F = 5.03 \times 10^{-7}N

so above is the gravitational force between car and the person

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An object, initially at rest, is subject to an acceleration of 34 m/s^2. How long will it take for that object to reach 3400m ?
Norma-Jean [14]
Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

a=v/t
t=v/a
t=480.83/34
t=14.142 s
6 0
3 years ago
3. A car has a mass of 2.50 x 10^3 kg. If the force acting on the car is 7.65 x 10^3 N to the
const2013 [10]

Answer:

3.06m/s² to the east

Explanation:

Given parameters:

Mass of car = 2.5 x 10³kg

Force acting on the car  = 7.65 x 10³N

Unknown:

Acceleration of the car  = ?

Solution:

From Newton's second law of motion:

      Force  = mass x acceleration

   Acceleration  = \frac{Force }{mass}   = \frac{7.65 x 10^{3} }{2.5 x 10^{3} }    = 3.06m/s² to the east

6 0
3 years ago
Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan
Drupady [299]

Answer:

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

Explanation:

We are given that

Gravitational force=F_g=\frac{Gm_Emr}{R^3_E}

r=0,U(0)=0

We know that

Gravitational potential energy=-\int F_gdr

U(r)=-\int\frac{Gm_Emr}{R^3_E}dr

U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C

Substitute r=0 ,U(0)=0

0=0+C

C=0

Substitute the value

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

4 0
3 years ago
A sample of an unknown substance has a mass of 0. 158 kg. If 2,510. 0 J of heat is required to heat the substance from 32. 0°C
guapka [62]

The specific heat of the unknown substance with a mass of 0.158kg is 0.5478 J/g°C

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

The specific heat capacity of a substance can be calculated using the following formula:

Q = m × c × ∆T

Where;

  • Q = quantity of heat absorbed (J)
  • c = specific heat capacity (4.18 J/g°C)
  • m = mass of substance
  • ∆T = change in temperature (°C)

According to this question, 2,510.0 J of heat is required to heat the 0.158kg substance from 32.0°C to 61.0°C. The specific heat capacity can be calculated:

2510 = 158 × c × (61°C - 32°C)

2510 = 4582c

c = 2510 ÷ 4582

c = 0.5478 J/g°C

Therefore, the specific heat capacity of the unknown substance that has a mass of 0.158 kg is 0.5478 J/g°C.

Learn more about specific heat capacity at: brainly.com/question/2530523

4 0
3 years ago
190 kg of water is to be raised by a water pump to a height of 25 meters from the bottom of a well in 60 seconds. What should be
ladessa [460]

Answer:

776.6 w

1.04 hp

Explanation:

given:

Mass, m = 190kg

height change, h = 25m

time elapsed, t = 60 s

acceleration due to gravity, g = 9.81 m/s²

Potential energy required raising 190 kg of water to a height of 25m

= mgh

= 190 x 9.81 x 25

= 46,597.5 J

Power required in 60 s

= Energy required ÷ time elapsed

= 46,597.5 ÷ 60

= 776.6 Watts  (Use conversion 1 W = 0.00134102 hp)

= 776.6 w x 0.00134102 hp/w

= 1.04 hp

6 0
3 years ago
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