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Lelu [443]
2 years ago
8

Multiple choice: Sunspots appear dark because

Physics
1 answer:
zepelin [54]2 years ago
3 0

Answer: C. They are regions in which strong magnetic fields make it difficult for fresh supplies of hot, ionized gas to reach the photosphere

Explanation:

Sunspots appear dark because they are regions in which strong magnetic fields make it difficult for fresh supplies of hot, ionized gas to reach the photosphere.

Sunspots are typically cooler than their surroundings. For example, the temperature of a large sunspot can be about 4,000 Kelvin which is lower than the temperature of the photosphere around it which is about 5,800 Kelvin.

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An astronaut has a mass of 74.0 kg. 1) how much would the astronaut weigh on mars where surface gravity is 38.0% of that on eart
NeX [460]
274.614N or 61.736lbs
3 0
2 years ago
Read 2 more answers
In a second experiment, you decide to connect a string which has length L from a pivot to the side of block A (which has width d
Salsk061 [2.6K]

Answer:

The answer is in the explanation

Explanation:

A)

i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.

ii) The force on friction acting on the blocks is proportional to their mass, since mass of block B is less than block A, the force of friction acting on block B is also less. Hence, one might argue that block B will go farther along the table before coming to rest.

B) The equation of motion for block A is

m_{A}\frac{\mathrm{d} v}{\mathrm{d} t} = -m_{A}g\nu_{s}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t} = -\nu_{s}g \quad (1)

Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get

v(t) = C-\nu_{s}gt \quad (2)

Here, C is the constant of integration, which can be determined by using the initial condition

v(t=0) = v_{0}\Rightarrow C = v_{0} \quad (3)

Hence

v(t) = v_{0} - \nu_{s}gt \quad (4)

Block A will stop when its velocity will become zero,i.e

0 = v_{0}-\nu_{s}gT\Rightarrow T = \frac{v_{0}}{\nu_{s}g} \quad (5)

Going back to equation (4), we can write it as

\frac{\mathrm{d} x}{\mathrm{d} t} = v_{0}-\nu_{s}gt\Rightarrow x(t) = v_{0}t-\nu_{s}g\frac{t^{2}}{2}+D \quad (6)

Here, x(t) is the distance travelled by the block and D is again a constant of integration which can be determined by imposing the initial condition

x(t=0) = 0\Rightarrow D = 0 \quad (7)

The distance travelled by block A before stopping is

x(t=T) = v_{0}T-\nu_{s}g\frac{T^{2}}{2} = v_{0}\frac{v_{0}}{\nu_{s}g}-\nu_{s}g\frac{v_{0}^{2}}{2\nu_{s}^{2}g^{2}} = \frac{v_{0}^{2}}{2\nu_{s}g} \quad (8)

C) We can see that the expression for the distance travelled for block A is independent of its mass, therefore if we do the calculation for block B we will get the same result. Hence the reasoning for Student A and Student B are both correct, the effect of having larger initial energy due to larger mass is cancelled out by the effect of larger frictional force due to larger mass.

D)

i) The block A is moving in a circle of radius L+\frac{d}{2} , centered at the pivot, this is the distance of pivot from the center of mass of the block (assuming the block has uniform mass density). Because of circular motion there must be a centripetal force acting on the block in the radial direction, that must be provided by the tension in the string. Hence

T = \frac{m_{A}v^{2}}{L+\frac{d}{2}} \quad (9)

The speed of the block decreases with time due to friction, hence the speed of the block is maximum at the beginning of the motion, therfore the maximum tension is

T_{max} = \frac{m_{A}v_{0}^{2}}{L+\frac{d}{2}} \quad (10)

ii) The forces acting on the block are

a) Tension: Acting in the radially inwards direction, hence it is always perpendicular to the velocity of the block, therefore it does not change the speed of the block.

b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.

The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).

8 0
3 years ago
A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa
Lesechka [4]
At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
U_i= \frac{1}{2} ka^2
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
K_i =0
And the total energy of the system is
E_i = U_i+K= \frac{1}{2}ka^2

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
U_f = 0
while the mass is moving at speed v, and therefore the kinetic energy is
K_f =  \frac{1}{2} mv^2
And the total energy is
E_f = U_f + K_f =  \frac{1}{2} mv^2

For the law of conservation of energy, the total energy must be conserved, therefore E_i = E_f. So we  can write
\frac{1}{2} ka^2 =  \frac{1}{2}mv^2
that we can solve to find an expression for v:
v= \sqrt{ \frac{ka^2}{m} }
6 0
3 years ago
A circular coil 14.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this locati
sp2606 [1]

Answer:

a)T = 2.9*10^{-5} N-m

b) north edge will rise up

Explanation:

torque on the coil is given as

T = NIABsin\theta

where N is number of loop =  9 loops

i is current = 7.80 A

-B -earth magnetic field = 5.00*10^{-5} T

A- area of circular coil

A = \frac{\pi d^{2}}{4}

A =\frac{\pi .14^{2}}{4}

A =0.015 m2

PUTITNG ALL VALUE TO GET TORQUE

T = 9*7.8*0.015*5*10^{-5} sin{90-56}

T = 2.9*10^{-5} N-m

b) north edge will rise up

3 0
3 years ago
Read 2 more answers
Found ones similar, but not with this question. Someone help?
agasfer [191]

The distance between the plates is 7.1 * 10^-7 m

<h3>What is the distance?</h3>

The capacitor is a device that is used to store charges. The capacitor always has a dielectric that is placed between the capacitors that separates them.

Given that;

C = εoA/d

C = capacitance

εo = permittivity of free space

A = cross sectional area

d = distance apart

Cd = εoA

d = εoA/C

d = 8.8 * 10^-12 * 8.89 * 10^-4/1.11 * 10^-8

d = 7.1 * 10^-7 m

Learn more about capacitor:brainly.com/question/17176550

#SPJ1

6 0
1 year ago
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