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3 years ago

Why are the coldest places on earth found near the poles

1 answer:

6 0

Polar regions do not receive direct sunlight during the winter months due to the tilt in the Earth's axis. Hence, polar regions can get very cold. Antarctica is the coldest place on Earth. The coldest places on Earth tend to be located near the poles. Hope this answers the question.

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A 115-turn circular coil of radius 2.71 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coi

tester [92]

Answer:

80.6 mV

Explanation:

Parameters given:

Number of turns, N = 115

Radius of coil, r = 2.71 cm = 0.0271m

Time taken, t = 0.133s

Initial magnetic field, Bin = 50.1 mT = 0.0501 T

Final magnetic field, Bfin = 90.5 mT = 0.0905 T

Induces EMF is given as:

EMF = [(Bfin - Bin) * N * A] / t

EMF = [(0.0905 - 0.0501) * 115 * pi * 0.0271²] / 0.133

EMF = (0.0404 * 115 * 3.142 * 0.0007344) / 0.133

EMF = 0.0806 V = 80.6 mV

3 0

3 years ago

What is the area called where the particles are spread apart?

Ostrovityanka [42]

Answer: Because of the longitudinal motion of the air particles, there are regions in the air where the air particles are compressed together and other regions where the air particles are spread apart. These regions are known as compressions and rarefactions respectively

Explanation:

4 0

3 years ago

a roller coaster moves on a certain section of it's track with an average speed of 13 m/s. how much distance does it cover in 5.

viva [34]

Speed=13m/s
Time=5.8s

$\\ \sf\longmapsto Speed=\dfrac{Distance}{Time}$

$\\ \sf\longmapsto Distance=Speed\times Time$

$\\ \sf\longmapsto Distance=13(5.8)$

$\\ \sf\longmapsto Distance=75.4m$

7 0

3 years ago

Read 2 more answers

Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v

wlad13 [49]

1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we call $v_f$ the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
$m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f$ (1)
where
$m_A=7 kg$ is the mass of ball A
$m_B=2 kg$ is the mass of ball B
$v_{Ai}=6 m/s$ is the initial velocity of ball A
$v_{Bi}=-12 m/s$ is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find $v_f$ , we find that the final velocity of the balls is
$v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s$
and the positive sign means the two balls are going to the right.

2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
$m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}$
$\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2$
where
$v_{fA}$ is the final velocity of ball A
$v_{fB}$ is the final velocity of ball B

If we solve simultaneously the two equations, we find:
$v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s$
$v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s$
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
$p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s$
and the total momentum after the collision is:
$p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s$

3 0

3 years ago

Which type of galaxy often contains the largest relative amount of gas and dust?

Vikki [24]

the answer is spiral galaxies

4 0

3 years ago