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BlackZzzverrR [31]
3 years ago
8

Red shift data shows that galaxies are O expanding O shrinking O moving away O moving closer

Physics
1 answer:
Anon25 [30]3 years ago
7 0

Answer:

Should be moving away

Explanation:

Red is a longer wavelength therefore further away. Wavelength is stretched out more and on the red end. I hope this is right. I decided to research and answer since you didn’t have other answers. Are you taking this on edg? I hope I helped!

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What data will the simulation provide about your design? You will also need a control (something you don’t change) that you can
Ksenya-84 [330]

Answer:

yes you will need a control

Explanation: all i know is that you need a control i don't know how to set it up lol sorry

6 0
2 years ago
A flywheel in the form of a uniformly thick disk of radius 1.33 m1.33 m has a mass of 70.6 kg70.6 kg and spins counterclockwise
ladessa [460]

Answer:

The constant torque required to stop the disk is 8.6 N-m in clockwise direction .

Explanation:

Let counterclockwise be positive direction and clockwise be negative direction .

Given

Radius of disk , r = 1.33 m

Mass of disc , m = 70.6 kg

Initial angular velocity , \omega_i =217 rpm

Final angular velocity , \omega_f =0\, rpm

Time taken to stop , t = 2.75 min

Let \alpha  be the angular acceleration

We know

\omega _f=\omega _i+\alpha t

=>0=217+2.75\alpha =>\alpha = -78.9\frac{rev}{min^{2}}

=>\alpha =-\frac{78.9\times 2\pi}{60\times 60}\frac{rad}{s^{2}}=-0.138 \frac{rad}{s^{2}}

Torque required to stop is given by

\tau =I\alpha

where  moment of inertia , I=\frac{mr^{2}}{2}=\frac{70.6\times 1.33^{2}}{2}kg.m^{2}=62.5 kg.m^{2}

=>\therefore \tau =-0.138\times 62.5\, N.m=-8.6\, N.m

Thus the constant torque required to stop the disk is 8.6 N-m in clockwise direction .

3 0
3 years ago
You did 200 joules of work lifting a 150-newton backpack. How high did you lift the backpack?
kaheart [24]
We know that:
W=Fs
200J=150N*s
s=200J/150N
s=1,33m
8 0
4 years ago
Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
Inessa05 [86]

Answer:

+1.11\mu C

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

Q_1 is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to Q_2 and Q_3 at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of Q_1.

Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

6 0
3 years ago
Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 50.0 mm , and t
givi [52]

Explanation:

Below is an attachment containing the solution.

4 0
4 years ago
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