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4 years ago

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Under most conditions, except ________ change, the resistance of an object is a constant and does not depend on the amount of th

e voltage or the amount of current passing through it.

1 answer:

Readme [11.4K]4 years ago

4 0

Answer:

Temperature change

Explanation:

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A small, 150 g cart is moving at 1.60 m/s on a frictionless track when it collides with a larger, 5.00 kg cart at rest. After th

NemiM [27]

Answer:

Mass of larger cart will be 0.0213 m /sec

Explanation:

We have given mass of small cart $m_1=150gram=0.150kg$

Initial velocity of small cart $v_{1i}=1.6m/sec$

Mass of larger block $m_2=5kg$

Initial velocity of larger cart $v_{2i}=0m/sec$

After the collision velocity of smaller cart $v_{1f}=-0.890m/sec$

Now according conservation of momentum

$m_1v_{1i}+m_2v_{2i}=m_1v{1f}+m_2v_{2f}$

$0.150\times 1.6+5\times 0=0.150\times -0.890+5\times v_{2f}$

$5\times v_{2f}=0.1065$

$v_{2f}=0.0213m/sec$

5 0

3 years ago

A satellite in earth orbit has a mass of 99 kg and is at an altitude of 2.02 106 m. (assume that u = 0 as r â â.) (a) what is th

Ierofanga [76]

(a) What is the potential energy: PE = -G * M * m/r

Where: M is the mass of the earth which is 5.98 * 10^24 kg.

m is the mass of the satellite.

r is the space from the center of the earth to the satellite

To conclude this distance add the radius of the earth to the altitude. Radius of the earth is 6.38 * 10^6 meters.

r = 6.38 * 10^6 + 2.02 * 10^6 = 8.38 * 10^6

PE = 6.67 * 10^-11 * 5.98 * 10^24 * 99/8.38 * 10^6 = 4.71240095 * 10^9 J

(b) magnitude of the gravitational force exerted by the Earth

Fg = G * M * m/r^2

Fg = 6.67 * 10^-11 * 5.98 * 10^24 * 99/(8.38 * 10^6)^2 = 562.3078873 N

(c) There are no other forces that the satellite exert on the Earth. So therefore, it is 0.

8 0

3 years ago

What can you infer about a wave with a long wavelength?

katovenus [111]

Answer:

It has a high frequency

Explanation:

3 0

3 years ago

Read 2 more answers

| The electric field 5.0 cm from a very long charged wire is (2000 N/C, toward the wire). What is the charge (in nC) on a 1.0-cm

Serggg [28]

Answer:

The charge is 0.056 nC.

Explanation:

Given that,

Electric field = 2000 N/C

Distance = 5.0 cm

We need to calculate the charge density

Using formula of charge density

$E=\dfrac{\lambda}{2\pi\times\epsilon_{0}r}$

$\lambda=2\pi\times\epsilon_{0}\times r\times E$

Put the value into the formula

$\lambda=2\pi\times8.85\times10^{-12}\times5.0\times10^{-2}\times2000$

$\lambda=5.56\times10^{-9}\ C/m$

We need to calculate the charge in 1.0 cm

Using formula of charge

$Charge = \lambda\times\text{length of segment}$

$Charge =5.56\times10^{-9}\times1.0\times10^{-2}$

$Charge=0.056\times10^{-9}\ C$

$Charge=0.056\ nC$

Hence, The charge is 0.056 nC.

3 0

3 years ago

**HELP ASAP PLEASE!**

Setler79 [48]

They use a spectrograph so the answer would be B.

7 0

3 years ago

Read 2 more answers