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3 years ago

What is the mass of a 735-N weight on Earth?

Physics

1 answer:

Allisa [31]3 years ago

3 0

W = mg, Assuming g ≈ 9.8 m/s² on the earth surface.

735 N = m* 9.8

735/9.8 = m

75 = m

Mass , m = 75 kg. B.

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How many significant figures should be written in the sum 4.65g +9.032g+580.0078+540.439g?

Ne4ueva [31]

Answer: 1.13 X 10^3 g or 1130 g

Explanation: When you add up everything, you will get 1134.1288 g = 1.11341218 X 10^3. But while adding significant figures, we always look for the least significant figures (here 4.65 has least sig. figures i.e. 3) and take that as a reference to how many significant figures we should have in the answer. (Pretty hard to explain!) Hope you got it!!

7 0

2 years ago

Which switch configuration creates a short circuit?

lara31 [8.8K]

I think the answer would be Switch 1 open; switches 2, 3, and 4 closed. It is letter D. This will happen when there is a low resistance and a high circuit so from the choices above, when two bulbs are open and one bulb is close, so there is a low resistance and high voltage.

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3 years ago

A disk-shaped merry-go-round of radius 2.63 m and mass 152 kg rotates freely with an angular speed of 0.526 rev/s. A 51.7 kg per

zalisa [80]

Explanation:

Given

radius $r=2.63 m$

$N=0.526 rev/s$

$\omega =3.30 rad/s$

mass disc $M=152 kg$

mass of person $m=51.7 kg$

velocity of Person $v=2.76 m/s$

moment of inertia $I=Mr^2$

$I=0.5\times 152\times 2.63^2=827.64 kg-m^2$

Initial angular momentum

$L_i=I\omega +mvr$

$L_i=827.64\times 3.30+51.7\times 2.76\times 2.63$

$L_i=2731.212+375.27=3106.48 Js$

Final Moment inertia

$I_f=0.5Mr^2+mr^2$

$I_f=(152\cdot 0.5+51.7)\cdot (2.63)^2=1185.243 kg-m^2$

final angular momentum

$L_f=I_f\omega _f$

Conserving angular momentum

$L_i=L_f$

$3106.48=1408.97\times \omega _f$

$\omega _f=2.62 rad/s$

4 0

3 years ago

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

sergij07 [2.7K]

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

Radius of the coil, r = 20 cm = 0.2 m

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 :

$\epsilon=NBA\omega$