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viktelen [127]
2 years ago
5

A galaxy that is a featureless spherical ball of stars would be called a type

Physics
1 answer:
satela [25.4K]2 years ago
7 0

Answer:

E0

Explanation:

Yes. The "0" indicates that it is spherical

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A positive test charge q is released from rest at distance r away from a charge of Q and a distance 2r away from a charge of 2Q.
Luba_88 [7]

Answer: Option (b) is the correct answer.

Explanation:

It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.

Then test charge to the right immediately after being released.

Therefore, the net force will be as follows.

            F = \frac{kqQ}{r^{2}} - kq\frac{(2Q)}{(2r)^{2}}

               = \frac{4KqQ - 2KqQ}{4r^{2}}

               = \frac{KqQ}{2r^{2}}

           F = \frac{KqQ}{2r^{2}} > 0

Thus, we can conclude that the test charge move to the right immediately after being released.

7 0
3 years ago
Flexibility is earned through regular _____ and ______
grin007 [14]

Answer:

D.

Explanation:

7 0
2 years ago
You set out to design a car that uses the energy stored in a flywheel consisting of a uniform 101-kg cylinder of radius r that h
Ket [755]
Ok, assuming "mj" in the question is Megajoules MJ) you need a total amount of rotational kinetic energy in the fly wheel at the beginning of the trip that equals
(2.4e6 J/km)x(300 km)=7.2e8 J
The expression for rotational kinetic energy is

E = (1/2)Iω²  

where I is the moment of inertia of the fly wheel and ω is the angular velocity.  
So this comes down to finding the value of I that gives the required energy.  We know the mass is 101kg.  The formula for a solid cylinder's moment of inertia is

 I = (1/2)mR²

We want (1/2)Iω² = 7.2e8 J and we know ω is limited to 470 revs/sec.  However, ω must be in radians per second so multiply it by 2π to get 
ω = 2953.1 rad/s
Now let's use this to solve the energy equation, E = (1/2)Iω²,  for I:
I = 2(7.2e8 J)/(2953.1 rad/s)² = 165.12 kg·m²

Now find the radius R,

 165.12 kg·m² = (1/2)(101)R²,
√(2·165/101) = 1.807m

R = 1.807m
8 0
3 years ago
A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le
dangina [55]

Momentum should be conserved. The momentum of both objects must balance with their initial and final momentum.

Let m1 and v1 be the mass and velocity of the bowling ball

And m2 and v2 be the mass and velocity of the bowling pin

(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f

30 kg m/s + (1.5 kg)(0 m/s) = 13kg m/s + 1.5v2f

V2f = 11.33 m/s

<span>So the momentum = 1.5 kg(11.33 m/s) = 17 kg m/s</span>

8 0
3 years ago
Bats use a process called echolocation to find their food. This involves giving out sound waves that hit possible prey or food.
cupoosta [38]
The type of waves used by bats are sound waves. Most of the species use their larynx to produce ultrasound waves in the frequency range of 20 to 200 kilohertz.
These sound waves are echoed, reflected, by surroundings, in this case food or prey. These reflections are received by the specialized receptor cells in the ears of bats. The reflections are analyzed by the brain to make an image.
Fun fact: The brain cells of bats are also specialized to better analyze the frequency of ultrasound used by the bat.
8 0
3 years ago
Read 2 more answers
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