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3 years ago

What is the gravitational force between Earth and a 10 kg object on Earth’s surface?

Physics

1 answer:

4vir4ik [10]3 years ago

6 0

Answer:

The force between the Earth and the object is, F = 9.61 x 10⁻¹² N

Explanation:

Given data,

The mass of the Earth, M = 5.9 x 10²⁴ kg

The mass of the object, m = 10 kg

The gravitational force between the earth and the object is given by the formula,

F = G Mm/R²

Substituting the given values in the above equation,

F = 6.673 x 10⁻¹¹ x 5.9 x 10²⁴ x 10 / 6.4 x 10⁶

= 9.61 x 10⁻¹² N

Hence, the force between the Earth and the object is, F = 9.61 x 10⁻¹² N

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An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an

kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁= m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

$v_{f1} = \sqrt{2*a_{1}*d_{1} }$ Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁ :Newton's second law

-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁ Equation (2)

∑Fy₁=0 : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁ = m*g*cos30°

We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m

-μ₁*g*cos30°+g*sin30° = a₁

g*(-μ₁*cos30°+sin30°) = a₁

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

$v_{f1} = \sqrt{2*3.2*3} }$

$v_{f1} =\sqrt{2*3.2*3}$

$v_{f1} = 4.38 m/s$

Rough surface kinematics

vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s

0 =4.38²+2*a₂*d₂ Equation (3)

Rough surface kinetics (μ= 0.3)

∑Fx₂=ma₂ :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂ Equation (4)

∑Fy₂= 0 :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂ We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0 =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= - 6.79*3 = 20.4 N*m

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0

4 years ago

An object floats in a beaker as shown. when it was put into the beaker, it displaced an amount of water into the graduated cylin

Tanzania [10]

Answer:

answer is C. 10 g

Explanation:

: When an object floats, it displaces an amount of water that has the same mass as itself. Therefore, the mass of the water in the graduated cylinder is equal to the mass of the object. We can see that there are 10 mL of water in the graduated cylinder. We also know that the density of water is 1 g/mL. Since each mL of water has a mass of 1 g, then 10 mL of water has a mass of 10 g. If the mass of the displaced water is 10 g, then the mass of the floating object is also 10 g.

6 0

3 years ago

Example of items that changed chemically

Ostrovityanka [42]

Burning of gases is one the example of chemical change

3 0

3 years ago

A ball of mass 8 kg falls from rest from a height of 100 m. Neglecting air resistance,calculate its kinetic energy after falling

astra-53 [7]

As an object falls from rest, its gravitational energy is converted to kinetic energy

G.P.E = K.E = mgh

K.E = (80 Kg)(9.8 m/s²)(30 m)

K.E. = 23,520 J

4 0

4 years ago

Read 2 more answers

The microwaves in a certain microwave oven have a wavelength of 12.2 cm. How wide must this oven be so that it will contain five

leva [86]

Answer:

$l = 0.305 \ m$

$f = 3.0*10^{11} \ Hz$

Explanation:

From the question we are told that

The wavelength is $\lambda = 12.2 \ cm = 0.122 \ m$

The number of antinodal planes of the electric field considered is n = 5

The width is mathematically represented as

$l = \frac{ n \lambda}{2}$

$l = \frac{5 * 0.122 }{ 2}$

$l = 0.305 \ m$

Generally the frequency the errors was made is mathematically represented as

$f = \frac{c}{\lamda_k}$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

$\lambda_k$ is the wavelength of the microwave has to be in order for there still to be five antinodal planes of the electric field along the width of the oven, which is mathematically represented as

$\lambda_k = \frac{ \lambda * \frac{0.04}{2} }{n/2}$

$\lambda_k = \frac{0.122*0.02}{5/2}$

$f = \frac{3.0*10^{8}}{0.000976}$

$f = 3.0*10^{11} \ Hz$

8 0

3 years ago