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LuckyWell [14K]
2 years ago
13

Which of the following best describes a spiral galaxy?

Physics
2 answers:
sergij07 [2.7K]2 years ago
8 0
Several billion stars arranged in arms spiraling outwards
elena55 [62]2 years ago
7 0
I think the answer is c 
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To practice problem-solving strategy 9.1 a strategy for conservation of momentum problems. an 80-kg quarterback jumps straight u
kherson [118]
Refer to the diagram shown below.

By definition momentum =  mass *  velocity.

Before throwing the ball:
The initial momentum is
P₁ = 0.

After throwing the ball:
Let u = the backward velocity of the quarterback.
The momentum is 
P₂ = (0.43 kg)*(15 m/s) + (80 kg)*(- u m/s)

Conservation of momentum requires that
P₂ = P₁
6.45 - 80u = 0
u = 6.45/80 = 0.0806 m/s

Answer: 0.08 m/s backward

6 0
3 years ago
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A person walks out of a store with a pillow at the top of an overflowing shopping cart. While walking to the car, the cart's whe
Mashutka [201]
The pillow will continue to go forward as Newton's first law states that an object will continue in it's path unless another force acts upon it
7 0
2 years ago
PLS HELP!!! The quarter back runs 75 meters down a football field in 12 seconds to score a touchdown. How fast did he run? A. 0.
atroni [7]

Answer:

B.6

Explanation:

speed = distance / time

= 75m / 12s

= 6.25 m/s

The closest answer is b. 6 m/s

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3 0
3 years ago
Compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite having a 0.25 volume fraction of fibers, ass
zmey [24]

Answer:

632.5 MPa

Explanation:

\sigma_{m} = Matrix stress at fiber failure = 10 MPa

V_f = Volume fraction of fiber = 0.25

\sigma_f = Fiber fracture strength = 2.5 GPa

The longitudinal strength of a composite is given by

\sigma_{cl}=\sigma_{m}(1-V_f)+\sigma_fV_f\\\Rightarrow \sigma_{cl}=10(1-0.25)+(2.5\times 10^3)\times 0.25\\\Rightarrow \sigma_{cl}=632.5\ MPa

The longitudinal strength of the aligned carbon fiber-epoxy matrix composite is 632.5 MPa

8 0
2 years ago
An initially stationary 4.3 kg object accelerates horizontally and uniformly to a speed of 11 m/s in 3.4 s. (a) In that 3.4 s in
Nataliya [291]

Answer:

The work done on the object by the force accelerating it is 520.31 J

Explanation:

Given;

mass of an object = 4.3 kg

horizontal velocity of the object, v = 11 m/s

time of acceleration, t = 3.4 s

work done is given as the product of force and distance

Work done = Fd

horizontal distance traveled by the object within 3.4 s, is calculated as follows;

X = Vt + ¹/₂gt², gravity has little or no influence on horizontal displacement, thus g = 0

X = Vt

X = 11*3.4 = 37.4 m

Force on the object, F = ma = m(v/t) = 4.3(11/3.4) = 13.912 N

work done = Fd = 13.912 x 37.4 = 520.31 J

Therefore, the work done on the object by the force accelerating it is 520.31 J

8 0
3 years ago
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