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Nimfa-mama [501]

2 years ago

13

How do I calculate the amount of work done using the information on the graphs given?

1 answer:

Nataly [62]2 years ago

3 0

Answer:with the power of math and physic

Explanation:

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Where might Michael look to find ideas about what the teacher thinks is important information to be learned for a test? a.Vocabu

masya89 [10]

D. B and C only
My sister just took the test and im sitting next to her and D is the correct answer

8 0

4 years ago

Read 2 more answers

A mosquito flies toward you with a velocity of 2.4 km/h [E]. If a distance of 35.0 m separates you and the mosquito initially, a

tangare [24]

First convert the speed of mosquito to m/s:

So the mosquito is flying at (2,400/3,600) m/s, or ⅔ m/s.

<span>

Since you are moving at 2m/s, so this makes the closing velocity between you and the mosquito to be 2⅔ m/s. </span>

Therefore the mosquito will hit your sunglasses at:<span>

35 m / (2⅔ m/s) = 13⅛ seconds.

2.0 m/s * 13⅛ s = 26¼ m from your initial position.

<span>⅔ m/s * 13⅛ s = 8¾ m from the mosquito's initial position. </span></span>

7 0

3 years ago

Explain how Copernicus concluded that stars were farther away than planets. Draw a diagram showing how this principle applies to

Rainbow [258]

<span>Copernicus decided this with more of an educated guess than anything. For example is when your standing right next to a plane it's huge Right? Well when it's flying it looks really small. He used the same reasoning for stars. Since it looks small it must be farther away.</span>

8 0

3 years ago

For a short time the missile moves along the parabolic path y=(18−2x2) km. If motion along the ground is measured as x=(4t−3) km

muminat

Answer with Explanation:

We are given that

$y=(18-2x^2) km$

$x=(4t-3)km$

Differentiate x and y w.r.t t

$\frac{dx}{dt}=4$

$\frac{dy}{dt}=-4x\frac{dx}{dt}$

$\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)$

$v_x=\frac{dx}{dt}=4$

$v_y=\frac{dy}{dt}=-16(4t-3)$

Substitute t=1

$v_x=4$

$v_y=-16(4-3)=-16$

Magnitude of velocity= $\mid v\mid=\sqrt{v^2_x+v^2_y}$

$\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s$

Hence, the magnitude of the missile's velocity=16.49 m/s

$a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0$

$a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64$

Substitute t=1

$a_x=0,a_y=-64$

$\mid a\mid=\sqrt{a^2_x+a^2_y}$

$\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2$

Hence, the magnitude of acceleration when t=1 s= $64m/s^2$

3 0

3 years ago

A projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure.

marin [14]

Answer: A

Explanation:

I want my points so yea

4 0

2 years ago