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3 years ago

9

An object weighs 1200 newtons on a planet. what is the gravitational field strength of this planet if rthe object is 60 kilogram

s

1 answer:

ratelena [41]3 years ago

3 0

G = F/m
= 1200/60
= 20 N/Kg

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If I wanted to generate a maximum emf of 20 V, what angular velocity (radians/sec, aka Hz) would be required given a circular co

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Answer:

Angular velocity, $\omega=35.36\ rad/s$

Explanation:

It is given that,

Maximum emf generated in the coil, $\epsilon=20\ V$

Diameter of the coil, d = 40 cm

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Number of turns in the coil, N = 500

Magnetic field in the coil, $B=9\times 10^{-3}\ T$

The angle between the area vector and the magnet field vector varies from 0 to 2 π radians. The formula for the maximum emf generated in the coil is given by :

$\epsilon=NBA\omega$

$\omega=\dfrac{\epsilon}{NBA}$

$\omega=\dfrac{20\ V}{500\times 9\times 10^{-3}\ T\times \pi (0.2\ m)^2}$

$\omega=35.36\ rad/s$

So, the angular velocity of the circular coil is 35.36 rad/s. Hence, this is the required solution.

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2 years ago

Which labels correctly identify the layers most closely associated with gamma rays and visible light? Z: Gamma rays X: Visible l

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3 years ago

A pole-vaulter just clears the bar at 5.31 m and falls back to the ground. The change in the vaulter's potential energy during t

irina [24]

Answer:

Weight = 734.46 N

Explanation:

Given:

Initial height of the pole-vaulter is, $h_i=5.31\ m$

Final height of the pole-vaulter is, $h_f=0\ m$

Change in the vaulter's potential energy is, $\Delta U=-3900\ J$

We know that, change in potential energy is given as:

$\Delta U=mg(h_f-h_i)$

Where, 'm' is the mass of the object, 'g' is acceleration due to gravity and has a value of 9.8 m/s².

Now, weight of the object is given as the product of its mass and acceleration due to gravity. So, replace 'mg' by weight 'w'. So, the equation becomes,

$\Delta U = w(h_f-h_i)$

Now, rewriting in terms of 'w', we get:

$w=\dfrac{\Delta U}{(h_f-h_i)}$

Now, plug in all the given values and solve for 'w'. This gives,

$w=\dfrac{-3900\ J}{(0-5.31)\ m}\\\\\\w=\dfrac{3900}{5.31}\ N\\\\\\w=734.46\ N$

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8 0

3 years ago