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kvv77 [185]
3 years ago
12

What is the definition of Newton and could you leave an example of it too?

Physics
1 answer:
vichka [17]3 years ago
7 0
What is the definition of newton
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The magnitude of the gravitational field strength near Earth's surface is represented by
Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

F = G\cdot \frac{M\cdot m}{r^{2}}

Where:

M - Mass of the planet Earth, measured in kilograms.

m - Mass of the person, measured in kilograms.

r - Radius of the Earth, measured in meters.

G - Gravitational constant, measured in \frac{m^{3}}{kg\cdot s^{2}}.

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

F = m \cdot g

Where:

m - Mass of the person, measured in kilograms.

g - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

g = \frac{G\cdot M}{r^{2}}

Given that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972 \times 10^{24}\,kg and r = 6.371 \times 10^{6}\,m, the magnitude of the gravitational field near Earth's surface is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}

g \approx 9.82\,\frac{m}{s^{2}}

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

4 0
3 years ago
A car is travelling to the right with a speed of 40 m/s when the driver slams on the brakes. The car skids for 4 s with constant
vazorg [7]

Answer:

10m/s^2

Explanation:

Given data

velocity= 40m/s

time= 4 seconds

Acceleration a =????

We know that

a= velocity/time

substitute

a=40/4

a= 10m/s^2

Hence the acceleration will be 10m/s^2

3 0
3 years ago
A Parachutist with a camera, with descending at a speed of 12.5m/s, releases, the camera at an altitude of 64.3m. What is the ma
jonny [76]

Given :

Initial velocity, u = 12.5 m/s.

Height of camera, h = 64.3 m.

Acceleration due to gravity, g = 9.8 m/s².

To Find :

How long does it take the camera to reach the ground.

Solution :

By equation of motion :

h = ut+\dfrac{gt^2}{2}

Putting all given values, we get :

12.5t+\dfrac{9.8t^2}{2}=64.3\\\\4.9t^2+12.5t=64.3

t = 2.56  and t = −5.116.

Since, time cannot be negative.

t = 2.56 s.

Therefore, time taken is 2.56 s.

Hence, this is the required solution.

7 0
3 years ago
The Sun delivers an average power of 1.575 W/m2 to the top of Neptune's atmosphere. Find the magnitudes of max and max for the e
FrozenT [24]

Answer:

1.1486813808\times 10^{-7}\ T

34.46 V/m

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

c = Speed of light = 3\times 10^8\ m/s

I = Intensity = 1.575 W/m²

The maximum magnetic field intensity is given by

B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T

The magnetic field intensity is 1.1486813808\times 10^{-7}\ T

The maximum electric field intensity is given by

E_m=B_m\times c\\\Rightarrow E_m=1.1486813808\times 10^{-7}\times 3\times 10^8\\\Rightarrow E_m=34.46\ V/m

The  electric field intensity is 34.46 V/m

8 0
3 years ago
Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
Ede4ka [16]

Answer:

386.2^{\circ}F

Explanation:

We are given that

P_1=200lbf/in^2

P_2=60lbf/in^2

v_1=200ft/s

v_2=1700ft/s

T_1=500^{\circ}F

Q=0

C_p=1BTU/lb^{\circ}F

We have to find the exit temperature.

By steady energy flow equation

h_1+v^2_1+Q=h_2+v^2_2

C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

Substitute the values

1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}

500+1.598=T_2+115.4

T_2=500+1.598-115.4

T_2=386.2^{\circ}F

7 0
3 years ago
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