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Sveta_85 [38]
3 years ago
6

Two charged balloons 2.1 * 10ȹ meters apart have a repulsive force of 1.3 * 10ȹ newtons. The charge on one balloon is 8.2 * 10

È¥ coulombs. What is the charge on the other balloon? (k = 9.0 * 10© newton·metersÆ/coulombÆ)

Physics
2 answers:
Elanso [62]3 years ago
5 0
F12=(8.98×10↑9 m/f)(1C)(-4C)/(4m)↑2=(8.98×10↑9 N·m↑2/c↑2)(-0.25C↑2/m↑2)~2.25×10↑9N     i hope thats right 

Viktor [21]3 years ago
3 0

Correct answer is A. I'm taking this test right now. :)

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How does the change in the temperature of the universe provide evidence for universe expansion that supports the Big Bang Theory
xeze [42]

Question:

1) The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.

2) The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

3) The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

4) The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.​

Answer:

The correct option is;

3) The Universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses

Explanation:

With the temperature measurement carried out using the CSIRO radio telescope, Astronomers have been able to determine a temperature difference in the universe from 5.08 Kelvin 7.2 billion light years away to 2.73 Kelvin in the Universe today, which is in support of the Big Bang theory that as the Universe expanded from a state of extreme temperature that cools down as the Universe expands or the cosmos disperses.

8 0
3 years ago
A 1,200 kg dragster, starting from rest, reaches a maximum velocity of 140m/s in 5 seconds. At the 5 second mark, the dragster d
SSSSS [86.1K]

Answer:

Drag or air resistance

Explanation:

The force of friction caused by a moving fluid is called drag.  When that fluid is air, it's also known as air resistance.

8 0
3 years ago
A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic
Elena L [17]

Answer:

the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

Explanation:

The torque is given by :

\bar {N} = \bar {m} * \bar {B}

where ;

m = 0.160 A.m²

B = 0.0800 T

θ = 35°

So the magnitude of the torque N = mBsinθ

N = (0.160)(0.0800)(sin 35°)

N = 0.007341

N = 7.34×10⁻³ Nm

Hence, the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

b) The potential energy \bar{U} = \bar{-m} * \bar{B}

U = -mBcosθ

U = (- 0.160)(0.0800)(cos 45)

U = -0.010485

U = -1.0485 ×10⁻² J

Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

6 0
3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
Two cars, initially at rest and 5 km apart at t=0 , simultaneously move toward each other. Car A travels at a constant speed of
Anastasy [175]

Answer:

<em>d. 268 s</em>

Explanation:

<u>Constant Speed Motion</u>

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

\displaystyle v=\frac{d}{t}

Where  

v = Speed of the object

d = Distance traveled

t  = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

\displaystyle t=\frac{d}{v}

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.

The time it will take for them to meet is:

\displaystyle t=\frac{5}{67}

t = 0.0746 hours

Converting to seconds: 0.0746*3600 = 268.56

The closest answer is d. 268 s

8 0
3 years ago
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