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IRINA_888 [86]
2 years ago
10

What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.

0001 N?
Physics
2 answers:
Sidana [21]2 years ago
5 0

Answer:

<em>The distance between the compact car and pickup truck is 0.96048 m</em>

Explanation:

The gravitational force is directly proportional to the product of the masses of the interacting object, it is also inversely proportional to the square of the distance between them.  This is shown in equation 1;

F =G \frac{m_{1} X m_{2}  }{d^{2} }............ 1

Where F is the gravitational force = 0.0001 N

G is the gravitational constant = 6.673 x 10^{-11} Nm^{2} kg^{-2}

m_{1}  is the mass of the compact car = 900kg

m_{2} is the mass of the pickup truck = 1600kg

d is the distance and its unknown ?

Let us make d the subject formula in equation 1

d = \sqrt{G\frac{m_{1} m_{2}  }{F  } } .... 2

Substituting into equation 2 we have

d = \sqrt{\frac{6.673x10^{-11} x  900 x 1600}{0.0001N} }

d = 0.96048m

Therefore the distance between the compact car and pickup truck is 0.96048 m

elena-14-01-66 [18.8K]2 years ago
3 0

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

r= \sqrt{\frac{Gm1m2}{F} }

r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

r= \sqrt{\frac{Gm1m2}{F} }

r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m

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The figure shows the arrangement of master cylinder and slave cylinder of a part of braking system. The master cylinder piston,
Neko [114]

Answer:

a) 35 kPa

d) 140 N

c) 1) Increasing the brake fluid  pressure

2)  Increasing the slave piston surface area.

Explanation:

The parameters given are;

a) Force applied to the master cylinder piston = 28 N

Cross sectional area of the master cylinder piston = 8 cm² = 8 × 10⁻⁴ m²

The pressure P on the brake fluid is given by the formula for pressure as follows;

P= \dfrac{Applied \ force}{Area \over \ which \  force \ is \ applied} = \dfrac{28 \, N}{8 \times 10^{-4} \, m^2} = 35,000 \, N/m^2 = 35,000 \, Pa

The pressure on the brake fluid, P, produced by the master cylinder piston = 35,000 Pa = 35 kPa

b) Given that the area of the slave piston = 40 cm² = 0.004 m², we have from the formula for pressure, P;

P= \dfrac{Applied \ force}{Area \over \ which \  force \ is \ applied} = \dfrac{Applied \ force}{4 \times 10^{-3} \, m^2} = 35,000 \, N/m^2

Therefore;

Applied force on the slave piston = 4 × 10⁻³ m² × 35,000 N/m² = 140 N

c) The force, F produced by the slave cylinder piston is given by the relation;

F = Pressure × Area

Therefore, the two ways of increasing the force produced by the slave cylinder piston is as follows;

1) Increasing the pressure in the brake fluid by increasing the force exerted by the master cylinder piston

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3 years ago
How much work do you need to do if you use a force of 3 Newtons to move a table 25 meters?
Harlamova29_29 [7]

The work needed if a force of 3N is used to move a table 25 metres is 75Nm (joules).

<h3>How to calculate work done?</h3>

Work is a measure of energy expended in moving an object. This means that no work is done if the object does not move.

Force is a physical quantity that denotes the ability to push, pull, twist or accelerate a body and which has a direction and is measured in Newtons.

Work done on an object can be calculated by multiplying the force applied to move the object by the distance the object moves as follows:

Work done = Force (N) × distance (m)

According to this question, a force of 3 Newtons is used to move a table over a distance of 25 meters. The work done on the table can be calculated as follows;

W = 3N × 25metres

W = 75Nm

Therefore, 75Nm is the work done on the table.

Learn more about work done at: brainly.com/question/2117638

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6 0
11 months ago
In a circus performance, a monkey on a sled is given an initial speed of 4.3 m/s up a 24◦ incline. The combined mass of the monk
Serggg [28]

Answer:

d = 1.15 m

Explanation:

  • In absence of friction, the change in kinetic energy of the combined mass of the monkey and the sled, must be equal (with opposite sign), to the change in gravitational potential energy:

        ΔK = -ΔU

  • When friction is not negligible, the change in mechanical energy, must be equal to the work done by non-conservative forces (kinetic friction in this case):

       ΔK + ΔU = Wnc (1)

  • As the monkey + sled reach to the maximum distance up the incline, they will come momentarily to a stop, so the final kinetic energy is 0.

        (K_{f} -K_{o}) = 0 - \frac{1}{2} * m*v_{o} ^{2} = -\frac{1}{2} *22.5kg*(4.3m/s)^{2} = -208.1J

  • The change in gravitational energy, can be written as follows:

        (U_{f} - U_{o} ) = m*g*h - 0 = m*g*h = \\ \\ 22.5 kg*9.8 m/s2*d*sin (24 deg) = 89.7J*d

  • The sum of these two quantities, must be equal to the work done by the friction force, along the distance d up the incline:

        W_{nc} = -\mu k*N*d

  • The normal force, always normal to the surface, must be equal and opposite to the component of the weight normal to the incline:

        N = m*g*cos \theta = m*g*cos (24 deg) = \\ \\ 22.5 kg*9.8m/s2*0.913 = 201.4 N

  • Replacing in the equation for Wnc:

        W_{nc} = -\mu k*N*d = -0.45*201.4 N*d = -90.6 N*d

  • We can return to the equation (1) and solve for d:

        -208.1 J + 89.7N*d = -90.6N*d\\\\  d = \frac{208.1}{180.3} =1.15 m

3 0
3 years ago
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