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PolarNik [594]
3 years ago
12

A rower in a boat pushes the water using an oar.

Physics
2 answers:
brilliants [131]3 years ago
6 0

Answer:

It's B, because of process of elimination:

A: The rower would be applying force to the oar not the other way around. So that's immediately out.

C: If the force of the oar applies to the water, the force of the water cannot apply to the oar.

D: Same as why A is incorrect.

So B is the only one left.

solong [7]3 years ago
6 0
The correct answer is B. “Every action has an equal or opposite reaction”.
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3 years ago
1) A pendulum is configured to have a period of 2 seconds.
vazorg [7]

"The" (and any subsequent words) was ignored because we limit queries to 32 words.

3 0
3 years ago
Which term means the distance between two bridge supports?
Ad libitum [116K]

The best answer is b - span.

A span is the distance between two bridge supports The supports may be towers, columns, or even the wall of a canyon.

There are many kinds of bridges  but they all fall into three types namely beam, arch and suspension. The major difference between these three kinds of bridges is the distance that each can cross in  a single span.

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5 0
3 years ago
In 0.60 seconds, a projectile goes from 0 to 610 m/s. What is the acceleration of the projectile?
IceJOKER [234]

Answer: a=1016.66 m/s^{2}

Explanation:

Acceleration a is expressed in the following formula:

a=\frac{V_{f}-V_{o}}{t}

Where:

V_{f}=610 m/s is the final velocity of the projectile

V_{o}=0 m/s  is the initial velocity of the projectile

t=0.6 s is the time

Solving:

a=\frac{610 m/s-0 m/s}{0.6 s}

a=1016.66 m/s^{2} This is the acceleration of the projectile

6 0
4 years ago
If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re
valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J

So, the required work done is 16 J.

7 0
3 years ago
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