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kotegsom [21]
3 years ago
5

Quinn accelerates her skateboard along a straight path from 0 m/s to 4.0 m/s in 2.5 seconds. Calculate the acceleration of the s

kateboard. Use the following formula:
Variable
Equation
Solve

Physics
2 answers:
Sedaia [141]3 years ago
7 0

Answer: 16 m/s^2

Explanation:

Vf = Vi + at

40 = 0 + 2.5a

a = 40 / 2.5 = 16 m/s^2

Gnom [1K]3 years ago
5 0

Answer:

that's my answer using the normal formula to calculate acceleration

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If he leaves the ramp with a speed of 35.0 m/s and has a speed of 33.0 m/s at the top of his trajectory, determine his maximum h
nadezda [96]

Answer:

H = 6.93 m

Explanation:

given data

velocity v = 35 m/s

horizontal component Vx = 33 m/s

solution

we get here maximum height so first we get vertical component here that is express as

Vy = \sqrt{v^2- Vx^2}        .........................1

put here value

Vy = \sqrt{35^2- 33^2}

Vy = 11.66 m/s

and

now we get height

H = \frac{Vy^2}{2g}        .............................2

put here value

H = \frac{11.66^2}{2\times 9.8}

H = 6.93 m

7 0
3 years ago
ASAP
scoundrel [369]

Answer:

A

Explanation:

Hooke's law! F(spring)=-kx

There's no tricky square law here. The spring constant doesn't change, only x (distance stretched) changes. Therefore, if distance is halved, Force will be halved.

5 0
3 years ago
1. ability to cause change * Pick One PLEASE HELP ME
kkurt [141]

Answer:

electromagnetic wave  i think

Explanation:

8 0
4 years ago
Read 2 more answers
A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 mm from the bulb, the light i
slega [8]

Answer:

The appropriate solution is:

(a) \frac{1}{4}(I_o)

(b) \frac{1}{4} (u_o)

(c) \frac{1}{2}B_o

Explanation:

According to the question, the value is:

Power of bulb,

= 60 W

Distance,

= 1.0 mm

Now,

(a)

⇒  \frac{I}{I_o} =\frac{r_o_2}{r_2}

On applying cross-multiplication, we get

⇒  I=I_o\times \frac{1_2}{2^2}

⇒     =I_o\times \frac{1}{4}

⇒     =\frac{1}{4} (I_o)

(b)

As we know,

⇒ \frac{u}{u_o} =\frac{I}{I_o}

By putting the values, we get

⇒ u=\frac{1}{4}(u_o)

(c)

⇒ \frac{B^2}{B_o^2} =\frac{u}{u_o}

         =\frac{I}{I_o}

⇒ B=B_o\times \sqrt{\frac{1}{4} }

⇒     =\frac{1}{2}(B_o)

4 0
3 years ago
Describe the motion above and answer
amm1812

Answer:

mmmmmmmm

Explanation:

mmmmmmmmmmmmmmmmmmmm

7 0
3 years ago
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