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kotegsom [21]
3 years ago
13

A rock is thrown horizontally from a bridge with a velocity of 17.0 m/s. It takes the rock 3.0 s to strike the water below. What

is the magnitude of the velocity of the rock just before it hits the water
Physics
1 answer:
zalisa [80]3 years ago
7 0

Answer:

V= 33.98 m/s

Explanation:

Given that

Horizontal speed ,u= 17 m/s

Time taken by rockets to strike the water ,t= 3 s

We know that acceleration due to gravity ,g= 9.81 m/s²

There is no any acceleration in the horizontal direction that is why the horizontal veloity will remain constant.

In the vertical direction

vy = uy+ g t

Initial velocity in vertical direction is 0 m/s.

vy= 0+ 9.81 x 3

vy = 29.43 m/s

The resultant velocity

V=\sqrt{v_y^2+u^2}

V=\sqrt{29.43^2+17^2}\ m/s

V= 33.98 m/s

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Define the Element of Art, value. If you were using a pencil, how would you create a darker value?
ankoles [38]

Elements of art are stylistic features that are included within an art piece to help the artist communicate. The seven most common elements include line, shape, texture, form, space, colour and value, with the additions of mark making, and materiality.

You would create a darker value if you shaded it, shading it makes it darker.

8 0
3 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
2 years ago
They realize there is a thin film of oil on the surface of the puddle. If the index of refraction of the oil is 1.81, and they o
Sphinxa [80]

Answer:

The right solution is "165.8 nm".

Explanation:

Given:

Index of refraction,

n = 1.81

Wavelength,

λ = 600 nm

We know that,

⇒ t=\frac{\lambda}{2\times n}

By putting the values, we get

      =\frac{600}{2\times 1.81}

      =165.8 \ nm

3 0
3 years ago
A person trying to lose weight should
drek231 [11]
This is a very wierd question but I will answer.It is A

Hope I helped
5 0
3 years ago
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tresset_1 [31]
Acceleration, a =  (v - u)/t

where v is the final velocity, u is the initial velocity, and t is the time.

This formula on a velocity time graph represents the slope of the graph.
 
7 0
3 years ago
Read 2 more answers
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