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nika2105 [10]
3 years ago
11

PLEASE HELP! Need this ASAP (in 10 minutes)

Physics
2 answers:
xxMikexx [17]3 years ago
6 0
The answer to your question here is A,


(I am dragging a very helpful piece of information from a website call “The Physics Classroom” for an explain took by the way so go check them out if you are still having trouble.)

From the paragraph I read, I could pull out “The data convincingly show that wave frequency does not affect wave speed.”

Thus concluding my answer of why the speed is unchanged.

9966 [12]3 years ago
4 0
I think it’s B


step-by-step explanation:
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What is the acceleration of a softball if it has a mass of 0.5 kg and hits the cathers glove with a force of 25n
ioda
Acceleration is found if we have the force and mass. 

With the following equation: F = ma, we can find the missing values. 

F = 25n
M = 0.5 kg
a = ?

a = f/m
a = 25/0.5
a = 50

a = 50 m/s

So, the acceleration is 50 m/s^2 
3 0
3 years ago
Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal grav
DochEvi [55]

Answer:

a) 3673469.39 seconds

b) 6.61×10¹⁴ m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0.12×3×10⁸ m/s

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

Equation of motion

v=u+at\\\Rightarrow 0.12\times 3\times 10^8=0+9.8t\\\Rightarrow t=\frac{0.12\times 3\times 10^8}{9.8}=3673469.39\ s

Time taken to reach 12% of light speed is 3673469.39 seconds

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{(0.12\times 3\times 10^8)^2-0^2}{2\times 9.8}\\\Rightarrow s=6.61 \times 10^{14}\ m

The distance it would have to travel is 6.61×10¹⁴ m

7 0
3 years ago
What is the mechanical advantage of the screw shown below? O A. 14.1 O B. 2 O C. 12.6 O D. 8.2.​
Vilka [71]

Answer: C. 12.6

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11.304/0.9= 12.56

3 0
3 years ago
Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of
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Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner(v_1)=3.1 m/s

speed of second runner(v_2)=4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner t=\frac{295.6}{4.65}=63.56 s

time required by first runner to reach finish line=\frac{250}{3.1}=80.64 s

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0
4 years ago
A force of 10N is required to stretch a spring from 20cm to 25cm. What is the spring constant in N/m2 Be careful of unit
kupik [55]

Answer:

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If we ASSUME that the spring is un-stretched at the zero cm position

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5 0
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