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3 years ago

11

The sunspot cycle is a pattern of solar activity where the average number of sunspots gradually _______________________________

and _______________________________.A. Stops, re-startsB. Increases, burstsC. Fades, dies outD. Increases, decreases

1 answer:

zzz [600]3 years ago

8 0

Answer:

option D

Explanation:

Sunspots are the spot that appears on the sun, this spot appears darker than the surrounding surface of the sun.

Sun magnetic field goes through a cycle and this cycle is called the Sunspot cycle. Every 11 years the magnetic field of the sun completely flips. This sunspot cycle affects activity on the surface of the sun.

Sunspot cycle is the pattern of solar activity where an average number of sunspot gradually increase and decrease.

Hence, the correct answer is option D

You might be interested in

In which material does the light wave have the larger wavelength?

pochemuha

Answer: in the smaller/ thinner pieces.

8 0

2 years ago

A tennis player swings her 1000 g racket with a speed of 11 m/s. She hits a 60 g tennis ball that was approaching her at a speed

shusha [124]

Answer:

- 3.72 Ns.

9.44 m/s

Explanation:

mass of racket, M = 1000 g = 1 kg

mass of ball, m = 60 g = 0.06 kg

initial velocity of racket, U = 11 m/s

initial velocity of ball, u = 18 m/s

final velocity of ball, v = - 44 m/s

Let the final velocity of the racket is V.

(a) Momentum is defined as the product of mass and velocity of the ball.

initial momentum of the ball = m x u = 0.06 x 18 = 1.08 Ns

Final momentum of the ball = m x v = 0.06 x (- 44) = - 2.64 Ns

Change in momentum of the ball = final momentum - initial momentum

= - 2.64 - 1.08 = - 3.72 Ns

Thus, the change in momentum of the ball is - 3.72 Ns.

(b) By use of conservation of momentum

initial momentum of racket and ball = final momentum of racket and ball

1 x 11 + 0.06 x 18 = 1 x V - 0.06 x 44

12.08 = V - 2.64

V = 9.44 m/s

Thus, the final velocity of the racket afetr the impact is 9.44 m/s .

3 0

4 years ago

Please help! I haven't really been able to understand this topic :/

Anettt [7]

Ok but y I thought it was upside down tho...

4 0

3 years ago

Read 2 more answers

| A T-ball with a mass of 0.6 kg travels in the

r-ruslan [8.4K]

Answer:

$|I|=6\ Kg.m/s$

$F=120\ N$

Explanation:

Impulse and Momentum

They are similar concepts since they deal with the dynamics of objects having their status of motion changed by the sudden application of a force. The momentum at a given initial time is computed as

$p_o=m.v_o$

When a force is applied, the speed changes to $v_1$ and the new momentum is

$p_1=m.v_1$

The change of momentum is

$\Delta p=p_1-p_0=m(v_1-v_o)$

The impulse is equal to the change of momentum of an object and it's defined as the average net force applied times the time it takes to change the object's motion

$I=F.t=\Delta p$

Part 1

The T-ball initially travels at 10 m/s and then suddenly it's stopped by the glove. The final speed is zero, so

$\Delta p=0.6\ Kg(0-10\ m/s)=-6\ Kg.m/s$

The impulse is

$I=\Delta p$

$I=-6\ Kg.m/s$

The magnitude is

$|I|=6\ Kg.m/s$

Part 2

The force can be computed from the formula

$I=F.t$

The direction of the impulse the T-ball receives is opposite to the direction of the force exerted by the ball on the glove, thus $I_b=6\ kg.m/s$

$\displaystyle F=\frac{I}{t}=\frac{6\ kg.m/s}{0.05\ s}$

$\boxed{F=120\ N}$

7 0

3 years ago

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.

Genrish500 [490]

<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

Horizontal distance he must clear = 22 meter

The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

We have equation of motion, $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when displacement = 1.5 meter.

$1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds$

So the car has to cover a distance of 22 meter in 2.119 seconds.

So minimum speed required = 22/0.553 = 39.78 m/s

Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 $m/s^2$

Substituting

$-1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0$

In case of horizontal motion

Horizontal speed of car = V cos 7 = 0.993V

So it has to travel 22 meter in t seconds

0.993Vt = 22, Vt = 22.155 m

Substituting in the equation $4.9t^2-0.122Vt-1.5=0$

We will get $4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds$

Speed required = 22.155/0.926 = 23.93 m/s

Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

7 0

3 years ago