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Cloud [144]
3 years ago
14

If you are given force and distance you can determine power if you know

Physics
2 answers:
weeeeeb [17]3 years ago
8 0
Yes, we can determine power if we know force and distance.
As power= Force x velocity
and velocity= distance/ time.
SVETLANKA909090 [29]3 years ago
7 0
Knowing the force and distance tells you the amount of work.

You can determine power (rate of work) if you also know the
length of time time that the force lasted over the distance.
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Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S
Vlada [557]

Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M_S = 1.99 × 10³⁰ kg

Mass of the neutron star

M_N = 2( M_S )

M_N = 2( 1.99 × 10³⁰ kg )

M_N = ( 3.98 × 10³⁰ kg )

Radius of neutron star R_N = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω_N.

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM_N = / R_N² = mR_Nω_N²

ω_N² = GM_N = / R_N³

ω_N = √(GM_N = / R_N³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω_N = √( (  6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω_N = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω_N = √ 120831133.3636777

ω_N = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s

5 0
3 years ago
What is the value of the normal force if the coefficient of kinetic friction is 0.22 and the kinetic frictional force is 40 newt
Salsk061 [2.6K]
1.8x10 sqaured newtons for all plato users

5 0
3 years ago
Read 2 more answers
A body has translatory motion if it moves along a
Ierofanga [76]
<span>A body has translatory motion if it moves along a: mcqs </span>
8 0
3 years ago
A 3 kg object is attached to a 1000 N/m spring. The spring is compressed 0.10 m and then the spring launches the object horizont
Viktor [21]

Answer:

1.826m/s

Explanation:

E=1/2*k*(∆L)^2

E=1/2*mV^2

6 0
2 years ago
A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the
Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sin\theta = m\lambda

Where,

w = width

\lambda =wavelength

m is an integer, m = 1, 2, 3...

We here know that as sin\theta as w are constant, then

\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}

We need to find w_2, then

w_2 = \frac{w_1}{\lambda_1}\lambda_2

Replacing with our values:

w_2 = \frac{4.4*10^{-6}}{487}496

w_2 = 4.48*10^{-6}m

Therefore the width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

3 0
3 years ago
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