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3 years ago

Unit of gravitational force

Physics

1 answer:

azamat3 years ago

3 0

Answer:

Newton

Explanation:

9.80665 × kgf = 1 Newton N

1 kilogram force (kgf) was the force of gravity, which is pushing a mass of 1 kg at one place in the world on the ground; calculated according to Newton's law force = mass × acceleration.

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A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.

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The answer is B) 490.

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3 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

lianna [129]

<h2>Answer:</h2>

<em><u>Velocity of throwing arrow = 43.13 m/s.</u></em>

<h2>Explanation:</h2>

In the question,

Let us say the height from which the arrow was shot = h

Distance traveled by the arrow in horizontal = 61 m

Angle made by the arrow with the ground = 2°

So,

From the <u>equations of the motion</u>,

$61 =u.t\\t=\frac{61}{u}$

Now,

Also,

Finally, the angle made is 2 degrees with the horizontal.

So,

Final horizontal velocity = v.cos20°

Final vertical velocity = v.sin20°

Now,

u = v.cos20° (No acceleration in horizontal)

Also,

$v=u+at\\vsin20=0+9.8(t)\\t=\frac{v.sin20}{9.8}$

So,

We can say that,

$\frac{v.sin20}{9.8}=\frac{61}{v.cos20}\\v^{2}.sin20.cos20=597.8\\v^{2}=1860.56\\v=43.13\,m/s$

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3 years ago

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3 years ago

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A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi

Mamont248 [21]

Answer:

$a=368.97\ m/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega=0$

Acceleration of the wheel, $\alpha =7\ rad/s^2$

Rotation, $\theta=14\ rotation=14\times 2\pi =87.96\ rad$

Let t is the time. Using second equation of kinematics can be calculated using time.

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s$

Let $\omega_f$ is the final angular velocity and a is the radial component of acceleration.

$\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s$

Radial component of acceleration,

$a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2$

So, the required acceleration on the edge of the wheel is $368.97\ m/s^2$ .

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Unit of gravitational force ​

Unit of gravitational force