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3 years ago

What is the force that opposes the movement of an object through water??

2 answers:

5 0

In air, you'd call it air resistance, pilots call it drag. In water, some call it
water resistance, or just plain drag.

Whatever it's called, it's the friction between the object and the fluid, plus
the force needed to push the fluid out of the way to let the object get through.

vodomira [7]3 years ago

3 0

<span>Drag is the force that opposes the motion of a body through fluids including water</span>

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3 years ago

If an atom has 34 protons and 40 neutrons, what is its atomic number?

Inga [223]

The atomic number is 34. (A)

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4 years ago

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3 years ago

A 2.5. Gram arrow enters a 100 g apple with a speed of 115 m/s. If the apple was originally at rest, then what speed will it hav

stich3 [128]

Answer:

The speed of the apple will be 2.81 m/s when the arrow enters it.

Explanation:

We can find the speed of the apple by conservation of linear momentum:

$p_{i} = p_{f}$

$m_{ap}v_{ap_{i}} + m_{a}v_{a_{i}} = m_{ap}v_{ap_{f}} + m_{a}v_{a_{f}}$

Where:

$m_{ap}$ is the mass of the apple = 100 g = 0.1 kg

$m_{a}$ is the mass of the arrow = 2.5 g = 0.0025 kg

$v_{ap_{i}}$ and $v_{ap_{f}}$ is the initial and final speed of the apple respectively

$v_{a_{i}}$ and $v_{a_{f}}$ is the initial and final speed of the arrow respectively

Since the apple was originally at rest ( $v_{ap_{i}}$ = 0) and knowing that $v_{a_{f}}$ = $v_{ap_{f}}$ when the arrow enters into the apple, we have:

$0 + 0.0025 kg*115 m/s = v(0.0025 kg + 0.1 kg)$

$v = 2.81 m/s$

Therefore, the speed of the apple will be 2.81 m/s when the arrow enters it.

I hope it helps you!

5 0

3 years ago

One side of a triangle is increasing at a rate of 5 cm/sec and a second side is increasing at a rate of 7 cm/sec. If the area of

Vanyuwa [196]

Answer:

The rate of angle is 26.25 rad/sec.

Explanation:

Given that,

First side of triangle a= 20 cm

Second side of triangle b= 50 cm

One side of a triangle is increasing at a rate = 5 cm/sec

Second side is increasing at a rate = 7 cm/s

Angle $\theta=\dfrac{\pi}{3}=60^{\circ}$

If the area of the triangle remains constant,

We need to calculate rate of angle

Using formula of area of triangle

$A=\dfrac{1}{2}ab\sin\theta$

On differentiating

$\dfrac{dA}{dt}=\dfrac{1}{2}ab\cos\theta\dfrac{d\theta}{dt}+\dfrac{1}{2}a\sin\theta\dfrac{db}{dt}+\dfrac{1}{2}b\sin\theta\dfrac{da}{dt}$

Put the value into the formula

$0=\dfrac{1}{2}\times20\times50\cos60\dfrac{d\theta}{dt}+\dfrac{1}{2}\times20\sin60\times7+\dfrac{1}{2}\times50\sin60\times5$

$\dfrac{d\theta}{dt}=\dfrac{35\sqrt{3}\times\dfrac{25}{2}\sqrt{3}\times5}{250}$

$\dfrac{d\theta}{dt}=26.25\ rad/sec$

Hence, The rate of angle is 26.25 rad/sec.

8 0

3 years ago