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2 years ago

Using this formula a = F/m What acceleration results from exerting a 125N force on a 0.65kg

Physics

1 answer:

Softa [21]2 years ago

5 0

Answer:

Acceleration = 192.3 m/s² (Approx.)

Explanation:

Given:

Force = 125 N

Mass of ball = 0.65 kg

Find:

Acceleration

Computation:

We know that;

Acceleration = Force / Mas

So,

Acceleration = 125 / 0.65

Acceleration = 192.3 m/s² (Approx.)

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The centre of mass of a metre rule is at the 50cm mark. state what is meant by Centre of mass

Tanya [424]

Answer:

Centre of mass of any body is a point where all mass of a body is supposed to be concentrated

it lies in geometrical centre....

3 0

2 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

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

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago

A particle moves along the x-axis so that its velocity at anytime is t greater than or equal to 0 is given by v(t)=1-sin(2pi t)

UkoKoshka [18]

A) We differentiate the expression for velocity to obtain an expression for acceleration:
v(t) = 1 - sin(2πt)
dv/dt = -2πcos(2πt)
a = -2πcos(2πt)

b) Any value of t can be plugged in as long as it is greater than or equal to 0.

c) we integrate the expression of velocity to find an expression for displacement:
∫v(t) dt = ∫ 1 - sin(2πt) dt
x(t) = t + cos(2πt)/2π + c
x(0) = 0
0 = = + cos(0)/2π + c
c = -1/2π
x(t) = t + cos(2πt)/2π -1/2π

5 0

3 years ago

Which statement describes steps involved in the production of hydroelectric power?

viktelen [127]

Answer:

Answer is D.......Falling water turns a turbine that helps generate electricity.

Explanation:

Hydropower plants capture the energy of falling water to generate electricity. A turbine converts the kinetic energy of falling water into mechanical energy. Then a generator converts the mechanical energy from the turbine into electrical energy.

4 0

2 years ago

Read 2 more answers

In a Young's double-slit experiment, 610-nm-wavelength light is sent through the slits. The intensity at an angle of 2.95° from

Sliva [168]

Answer:

spacing between the slits is 405.32043 × $10^{-9}$ m

Explanation:

Given data

wavelength = 610 nm

angle = 2.95°

central bright fringe = 85%

to find out

spacing between the slits

solution

we know that spacing between slit is

I = 4 $I_{0}$ × cos²∅/2

I/4 $I_{0}$ = cos²∅/2

here I/4 $I_{0}$ is 85 % = 0.85

0.85 = cos²∅/2

cos∅/2 = √0.85

∅ = 2 × $cos^{-1}$ 0.921954

∅ = 45.56°

∅ = 45.56° ×π/180 = 0.7949 rad

and we know that here

∅ = 2π d sinθ / wavelength

d = ∅× wavelength / ( 2π sinθ )

put all value

d = 0.795 × 610× $10^{-9}$ / ( 2π sin2.95 )

d = 405.32043 × $10^{-9}$ m

spacing between the slits is 405.32043 × $10^{-9}$ m

7 0

2 years ago