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3 years ago

50 POINTS

Physics

2 answers:

mylen [45]3 years ago

8 0

Answer:

The correct answer is: B area = area

Explanation:

Vikki [24]3 years ago

3 0

Answer:

The correct answer is: B area = area

Explanation: You can solve this problem through the use of laws of Kepler's planetary motion. There are Kepler the laws ( 3 of them) In this exercise, we will use the second law. According to this law, a line segment joining a planet and the sun sweeps out areas is equal to that during the equal intervals of time

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suppose you want to determine the surface area of this sugar cube. it has edges that are each 2 cm long. if you cut the cube in

ivolga24 [154]

Answer:

Half: 6 cm^2 Whole: 12 cm^2

Explanation:

First, we know that the edges of the cube are 2 cm long. So there are 6 faces on a cube. We do 2x6=12 cm^2 as our total surface area. Then it asks for each half. So you would divide it by 2 and get 6 cm^2 as your half.

5 0

3 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

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

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

3 years ago

Jonathan must lift a 3000n rock using a lever. jonathan uses 300n of force to push down on the lever to move the rock. what is t

-Dominant- [34]

The load is the weight of the rock that Jonathan lifts:
$L=3000 N$
The effort instead is the force applied in input to the lever in order to lift the rock:
$E=300 N$

So, the ratio between load and effort for this exercise is
$\frac{L}{E}= \frac{3000 N}{300 N}=10$
So, the ratio is 10:1.

3 0

3 years ago

Where do you feel that you are traveling at the fastest speed when on the swing?

il63 [147K]

Answer:

Explanation:

I think it's C, because at that point, you are going fastest. Sorry if im wrong, hope this helps.

7 0

3 years ago

Read 2 more answers

Assume that a vaulter is able to carry a vaulting pole while running as fast

rewona [7]