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Serjik [45]
3 years ago
13

If Rojelio starts out at rest, and in 10 s speeds up to 50 m/s, what is his acceleration?

Physics
1 answer:
vovikov84 [41]3 years ago
3 0

Answer:

C) - 5m/s^2

Explanation:

U=0(started from rest)

V=50m/s

T=10s

Acceleration =v-u/t

=50-0/10

=5m/s^2

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If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?
Akimi4 [234]

Answer:

Bow Line

Explanation:

If the wind or current is pushing your boat away from the dock, bow line should be secured first.

1- We should cast off the bow and stern lines.

2-With the help of an oar or boat hook, keep the boat clear of the dock.

3-Leave the boat on its own for sometime and let the wind or current carry the boat away from the dock.

4 - As you see there is sufficient clearance, shift into forward gear and slowly leave the area.

7 0
2 years ago
1/A ball is dropped from the top of a building. After 3 seconds, its speed is measured to be 29.4 m/s. Calculate the acceleratio
AleksAgata [21]

Answer:

9.8

Explanation:

7 0
2 years ago
Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pull
bearhunter [10]

Answer:

strain = 1.4 \times 10^{-3}

Explanation:

As we know by the formula of elasticity that

E = \frac{stress}{strain}

now we have

E = 110 GPA

E = 110 \times 10^9 Pa

Area = 15.2 mm x 19.1 mm

A = 290.3 \times 10^{-6}

now we also know that force is given as

F = 44500 N

here we have

stress = Force / Area

stress = \frac{44500}{290.3 \times 10^{-6}}

stress = 1.53 \times 10^8 N/m^2

now from above formula we have

strain = \frac{stress}{E}

strain = \frac{1.53 \times 10^8}{110 \times 10^9}

strain = 1.4 \times 10^{-3}

7 0
3 years ago
Can someone please label this ?
Vadim26 [7]

Answer:

Explanation:

this is what i know so far ( some might be incorrect ) , but hope some of this helps!

3 0
3 years ago
A 69-kg student sitting on a hardwood floor does not slide until pulled by a 260-N horizontal force. find coefficient of frictio
salantis [7]

Answer:

μ = 0.385

Explanation:

Given that,

The mass of the student, m = 69 Kg

The horizontal force applied, F = 260 N

The normal force acting on the body, weight = mg = 69g  N

                                                                                    = 676.2 N

The coefficient of friction acting on a body is equal to the force acting on the body to the normal force acting on the body due to gravitation.

The formula for coefficient of friction,

                                    μ =  F / N

Substituting the values in the above equation,

                                     μ = 260 N / 676.2 N

                                        = 0.385

Hence, the  coefficient of friction, μ = 0.385

6 0
3 years ago
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