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3 years ago

In a tug of war, when one team is pulling with a force of 85 N and the other 40 N, what is the net

Physics

1 answer:

olya-2409 [2.1K]3 years ago

4 0

85 N - 40 N = 45 N
And depending on direction the greater force is being pulled towards

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If Spacecraft X has twice the mass of Spacecraft Y , then what is true about X and Y ?

Solnce55 [7]

I don't know what you were trying to copy or type in Choice-E .

All three statements are true, so the answer should say "I, II, and III".

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2 years ago

A transformer connected to a 120-V (rms) ac line is to supply 12,500 V (rms) for a neon sign. To reduce shock hazard, a fuse is

Tamiku [17]

Answer:

a) 104

b) 106

c) 884 mA

Explanation:

The ratio of the transformer is given by:

$N=\frac{V_{out}}{V_{in}}=\frac{12500V}{120V}=104$

We need to know the current in the primary in order to obtain the power applied.

$I_1=I_2*N=8.50mA*104=884mA\\\\P=I*V=884*10^{-3}A*120V=106W\\$

The current rating of the fuse is the current on the primary, 884mA as we calculated before in order to obtain the power.

4 0

3 years ago

An object moving 20 m/s

yan [13]

Answer:

We are given:

initial velocity (u) = 20m/s

acceleration (a) = 4 m/s²

time (t) = 8 seconds

displacement (s) = s m

Solving for Displacement:

From the seconds equation of motion:

s = ut + 1/2 * at²

replacing the variables

s = 20(8) + 1/2 * (4)*(8)*(8)

s = 160 + 128

s = 288 m

7 0

3 years ago

1. ________________electricity is the type of electricity commonly used in homes and businesses throughout the world.

amid [387]

Number 1 is letter A

3 0

3 years ago

A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m

elena-14-01-66 [18.8K]

Answer:

1,230N

Explanation:

1. Name of the variables:

$f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration$

2. Formulae:

$f=\dfrac{number\text{ }of\text{ }revolutions}{time}$

$\omega=2\pi f$

$a_c=\omega^2 r$

$F_c=m\times a_c$

3. Solution (calculations)

$f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}$

$\omega=2\pi\times0.\overline{60}\approx 3.808rad/s$

$a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2$

$F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N$

3 0

3 years ago