All categories

3 years ago

State Newton law of gravity

Physics

1 answer:

Nadusha1986 [10]3 years ago

8 0

Answer:

Newton's law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.

Explanation:

You might be interested in

A certain rain cloud at an altitude of 1.80 km contains 3.20 107 kg of water vapor. How long would it take for a 2.90-kW pump to

kvasek [131]

Answer:

2255 days

Explanation:

height, h = 1.8 km = 1800 m

amount of water, m = 3.2 x 10^7 kg

Power, P = 2.9 kW = 2900 W

Let t be the time taken

Energy required to lift the water,

E = m g h

E = 3.2 x 10^7 x 9.8 x 1800 = 5.65 x 10^11 J

Power, P = Energy / time

t = E / P = (5.65 x 10^11) / 2900

t = 1.95 x 10^8 second

t = 2255 days

thus, the time taken is 2255 days.

7 0

4 years ago

An electron is released from a negatively charged plate 0.5m away.The potential difference between the plates is 25v.

Citrus2011 [14]

Explanation:

In brief, electrons are negative charges and protons are positive charges. An electron is considered the smallest quantity of negative charge and a proton the smallest quantity of positive charge.

Two negative charges repel. Also, two positive charges repel. A positive charge and a negative charge attract each other (all experimentally verified.)

Point Charge: An accumulation of electric charges at a point (a tiny volume in space) is called a point charge.

Note: When an atom loses an electron, the separated electron forms a negative charge, but the remaining that contains one less electron or consequently one more proton becomes a positive charge. A positive charge is not necessarily a single proton. In most cases, a positive charge is an atom that has lost one or more electron(s).

6 0

3 years ago

Two blocks of weight 3.0N and 7.0N are connected by a massless string and slide down a 30 degree inclined plane. The coefficient

Luba_88 [7]

Answer:

a. a = $6.41 m/s^2$

b. T = -0.81 N

Explanation:

Given,

weight of the lighter block = $w_1\ =\ 3.0\ N$
weight of the heavier block = $w_2\ =\ 7.0\ N$
inclination angle = $\theta\ =\ 30^o$
coefficient of kinetic friction between the lighter block and the surface = $\mu_1\ =\ 0.13$

coefficient of kinetic friction between the heavier block and the surface = $\mu_2\ =\ 0.31$
friction force on the lighter block = $f_1\ =\ \mu_1N_1\ =\ \mu_1 w_1cos\theta$
friction force on the heavier block = $f_2\ =\ \mu_2N_2\ =\ \mu_2w_2cos\theta$

Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.

From the f.b.d. of the lighter block,

$w_1sin\theta\ -\ T\ -\ f_1\ =\ \dfrac{w_1a}{g}\\\Rightarrow T\ =\ w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

From the f.b.d. of the heavier block,

$w_2sin\theta\ +\ T\ -\ f_2\ =\ \dfrac{w_2a}{g}\\\Rightarrow T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)$

From eqn (1) and (2), we get,

$w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\\\Rightarrow w_1gsin\theta \ -\ w_1a\ -\ \mu_1w_1gcos\theta\ =\ w_2a\ -\ w_2gsin\theta\ -\ \mu_2 w_2gcos\theta\\$

$\Rightarrow a\ =\ \dfrac{g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta)}{w_1\ +\ w_2}\\\Rightarrow a\ =\ \dfrac{g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)]}{w_1\ +\ w_2}\\$

$\Rightarrow a\ =\ \dfrac{9.81\times [sin30^o\times (3.0\ +\ 7.0)\ +\ cos30^o\times (0.31\times 7.0\ -\ 0.13\times 3.0)]}{3.0\ +\ 7.0}\\\Rightarrow a\ =\ 6.4\ m/s.$

part (b)

From the eqn (2), we get, $T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ \mu_2w_2cos\theta\\\Rightarrow T\ =\ \dfrac{7.0\times 6.41}{9.81}\ -\ 7.0\times sin30^o\ -\ 0.31\times 7.0\times cos30^o\\\Rightarrow T\ =\ -0.81\ N$

3 0

3 years ago

Read 2 more answers

A car of mass 800 kg is travelling at 30 m/S. The car then brakes suddenly and stops with a braking distance of 75 m find the wo

gulaghasi [49]

Answer:

-360 kJ

Explanation:

Given:

m = 800 kg

v₀ = 30 m/s

v = 0 m/s

Δx = 75 m

Find: W

We can solve this using either forces or energy.

To use forces, first find the acceleration.

v² = v₀² + 2aΔx

(0 m/s)² = (30 m/s)² + 2a (75 m)

a = -6 m/s²

Then apply Newton's second law:

∑F = ma

F = (800 kg) (-6 m/s²)

F = -4800 N

Work is force times distance:

W = FΔx

W = (-4800 N) (75 m)

W = -360,000 J

W = -360 kJ

If you want to use energy instead:

work = change in energy

W = ΔKE

W = ½mv² − ½mv₀²

W = ½ (800 kg) (0 m/s)² − ½ (800 kg) (30 m/s)²

W = -360,000 J

W = -360 kJ

4 0

3 years ago

SOMEBODY PLEEASEEE HELP A STRUGGLING HIGHSCHOOLERRRR +(

CaHeK987 [17]

Explanation:

What is the weight of a 2.00-kilogram object on the surface of Earth?

2.00 N

4.91 N

9.81 N

19.6 N

Given parameters:

Mass of the object = 2kg

Unknown:

Weight of the object = ?

Solution:

The weight of an object is the force of gravity acting on the object;

Weight = mass x acceleration due to gravity

Acceleration due to gravity = 9.8m/s²

Now insert the parameters and solve;

Weight = 2 x 9.8 = 19.6N

A person weighing 785 Newtons on the surface of the Earth would weigh 47 Newtons on the surface of Pluto. What is the magnitude of the gravitational acceleration on the surface of Pluto?

1.7 m/s²

0.59 m/s²

0.29 m/s²

9.8 m/s²

Given parameters:

Weight on Earth = 785N

Weight on Pluto = 47N

Unknown:

Acceleration due to gravity on Pluto = ?

Solution

The mass of the body both on Earth and Pluto is the same.

Weight = mass x acceleration due to gravity

Now find the mass on Earth;

Acceleration due to gravity on Earth = 9.8m/s²

785 = mass x 9.8

mass = $\frac{785}{9.8}$ = 80.1kg

So;

Acceleration due to gravity on Pluto = $\frac{Weight on Pluto}{mass }$

Acceleration due to gravity = $\frac{47}{80.1}$ = 0.59m/s²

5 0

3 years ago

State Newton law of gravity​

State Newton law of gravity