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2 years ago

6

Help its due in 2 hours

1 answer:

mars1129 [50]2 years ago

5 0

Answer:

15.6 degrees

Explanation:

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A positively charged plastic ruler is brought close to a piece paper resting on the desk. The piece of paper was initially neutr

stealth61 [152]

Answer: static electricity

Explanation:

When the plastic ruler is rubbed, friction opposes the motion and causes the transfer of electron from one surface to another such that plastic becomes negatively charged. When ruler is brought nearer to the paper, it induces the positive charge in the piece of paper.

4 0

3 years ago

Consider a hot-air balloon. The deflated balloon, gondola, and two passengers have a combined mass of 315 kg. When inflated, the

vladimir2022 [97]

This will help you.

3 0

3 years ago

What is cheetah speed and strength?

pav-90 [236]

Answer:

Speed: 109.4–120.7 km/h (68.0–75.0 mph)

Strength: Couldn't find out.

Explanation:

3 0

2 years ago

The submarine emits a pulse of sound to detect other objects in the sea. The sped of sound in sea water is 1500m/s. An echo is r

r-ruslan [8.4K]

Answer:

d = 375 m

Explanation:

The speed of sound is constant in any medium, therefore we can use the uniform motion relationships

v = x / t

x = v t

In this case it indicates that the time since the sound is emitted and received is t = 0.50 s, in this time the sound traveled a round trip distance

x = 2d

2d = v t

d = v t/2

let's calculate

d = 1500 0.5 / 2

d = 375 m

3 0

3 years ago

Two long straight parallel wires are separated by 7.0cm. There is a 2.0A current flowing in the first wire. If the magnetic fiel

aliina [53]

Answer:

The current in the second wire is 4.4 A.

Explanation:

Given that,

Distance =7.0 cm

Current in first wire = 2.0 A

The magnetic field strength is zero at distance of 2.2 cm from the first wire.

We need to calculate the current in the second wire

Using formula of magnetic field

$B=B_{1}-B_{2}$

$0=\dfrac{\mu_{0}I_{1}}{2\pi r_{1}}-\dfrac{\mu_{0}I_{2}}{2\pi r_{2}}$

$\dfrac{\mu_{0}I_{1}}{2\pi r_{1}}=\dfrac{\mu_{0}I_{2}}{2\pi r_{2}}$

$\dfrac{I_{1}}{r_{1}}=\dfrac{I_{2}}{(d-r_{1})}$

Here, $r_{2}=d-r_{1}$

$I_{2}=\dfrac{I_{1}\times(d-r_{1})}{r_{1}}$

Put the value into the formula

$I_{2}=\dfrac{2.0\times(7.0-2.2)}{2.2}$

$I_{2}=4.4\ A$

Hence, The current in the second wire is 4.4 A.

4 0

3 years ago