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DedPeter [7]
3 years ago
5

Which of the preceding statements is/are true?A) Helioseismology is the study of the differential rotation and magnetic field of

the Sun. B) A filtergram is used to study layers below the photosphere. C) The chromosphere of the Sun has a higher temperature than the photosphere.
Physics
1 answer:
Nutka1998 [239]3 years ago
8 0

Answer:

A and C

Explanation:

Helioseismology is a branch of science that makes use of  differential rotation and magnetic field to study he structure of the sun.

A filtergram represents the photo of the surface of the sun that is made using high energy particles being accelerated by the sun's magnetic field. <em>Hence, it can only be used to study the surface and not the layers below the photosphere.</em>

The chromosphere of the sun has a temperature between 6,000 °C to about 20,000°C while the photosphere has a temperature between 4,230 and 5,730 °C.<em> Hence, the chromosphere has a higher temperature.</em>

The only true options are A and C.

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represents the space-time, speed-time and acceleration-time graphs for a ball pulled upwards with a speed of 10 m / s from 1 met
alexandr1967 [171]
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5 0
3 years ago
Consider an object with weight on the Earth W_earth. The gravity of the Earth is g. If it is moved to another planet with gravit
Mazyrski [523]

Answer:

W_{Planet} = \frac{W_{Earth}}{g}\times a

Explanation:

We know that weight of an object on Earth is,

W_{Earth} = m\times g

Thus,

m = \frac{W_{Earth}}{g}

where,

m = mass of an object, which is constant and is independent of gravity

g = acceleration due to gravity on Earth

On the new planet, gravity = a

Thus the weight of the object on the new planet will be

W_{Planet} = m\times a

W_{Planet} = \frac{W_{Earth}}{g}\times a

5 0
3 years ago
A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0
Xelga [282]

Answer:

v_m \approx -4.38\; \rm m \cdot s^{-1} (moving toward the incline.)

v_M \approx 4.02\; \rm m \cdot s^{-1} (moving away from the incline.)

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

If g = 9.81\; \rm m \cdot s^{-2}, the potential energy of the block of m = 2.20\; \rm kg would be m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this m = 2.20\; \rm kg\! block right before the collision would also be approximately 77.695\; \rm J.

Calculate the velocity of that m = 2.20\; \rm kg based on its kinetic energy:

\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}.

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right before the collision: approximately 18.489\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right after the collision: (m\cdot v_m + m \cdot v_M).

For momentum to conserve in this collision, v_m and v_M should ensure that m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

Kinetic energy of the two blocks right before the collision: approximately 77.695\; \rm J and 0\; \rm J. Sum of these two values: approximately 77.695\; \rm J\!.

Sum of the energy of each block right after the collision:

\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right).

Similarly, for kinetic energy to conserve in this collision, v_m and v_M should ensure that \displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J.

Combine to obtain two equations about v_m and v_M (given that m = 2.20\; \rm kg whereas M = 7.00\; \rm kg.)

\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right..

Solve for v_m and v_M (ignore the root where v_M = 0.)

\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right..

The collision flipped the sign of the velocity of the m = 2.20\; \rm kg block. In other words, this block is moving backwards towards the incline after the collision.

6 0
3 years ago
I NEED HELP!!!!! PLEASE ANSWER! FIRST CORRECT ANSWER WILL GET BRAINLIEST AND I WILL FOLLOW YOU!
Naily [24]

Average speed = (total distance) / (time to cover the distance)

We know:

    Average speed = 65 km/hr
    Total distance = 1,000 km 
    Time to cover it = (Driving Time) + 4 hours.

so we can write:

          65 km/hr  =  (1,000 km) / (Driving Time + 4hr)

          (I'm going to start calling the driving time 'DT'.
           Notice that DT is a number with the units of 'hours'.)

Multiply each side by    (DT + 4hr)

           (65 km/hr) (DT + 4hr)  =  1,000 km   

Eliminate parentheses on the left side:

           (65·DT km  +  260 km)  =  1,000 km

Subtract  260km  from each side:

              65·DT km          =    740 km

Divide each side by 65 :

                DT   =   11.38 hours .

DT (Driving Time) is the time you spent actually driving.
You had to cover the complete 1,000 km in that time.
So while you were driving, you had to do it at a speed of

                  1,000 km / 11.38 hrs  =  87.8 km/hr .
__________________________________________

As long as we're already totally bored by this question,
let's work on it some more, and check my answer:

... Driving for 11.38 hours at a speed of 87.8 km/hr, you cover

                     (11.38 hr) x (87.8 km/hr)  =  999.164 km  (close enough to 1,000) .

So far, so good.  The distance is taken care of.

With the 4-hour stop, the total trip takes 4 more hours = 15.38 hours.
So the average speed is

                     (1,000 km) / (15.38 hr)  =   65.02 km/hr

                                                                Close enough to 65 km/hr.  yay !

6 0
4 years ago
A Top Fuel Dragster initially at rest undergoes an average acceleration of 39 m/s2 for 3.97 seconds. How far will it go in this
tigry1 [53]

Answer:

s = 307.34 m

Explanation:

In order to find the distance covered by the dragster during the given time, we will use second equation of motion. The second equation of motion is written as follows:

s = Vi t + (0.5)at²

where,

s = distance covered by the dragster = ?

Vi = Initial Velocity = 0 m/s

t = time interval = 3.97 s

a = acceleration = 39 m/s²

Therefore,

s = (0 m/s)(3.97 s) + (0.5)(39 m/s²)(3.97 s)²

<u>s = 307.34 m</u>

5 0
3 years ago
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