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Paul [167]
3 years ago
12

An astronaut on the moon pushes with her left hand on a small rock and with her right hand on a large rock. Each hand pushes wit

h a force of 15 N. The small rock has mass 2 kg and the large rock has mass 8 kg. What forces do the rocks exert on the astronaut?
The small rock exerts a pushing force of 20 N; the large rock exerts a pushing force of 78 N.

The small rock exerts a pushing force of 3 N; the large rock exerts a pushing force of 12 N.

Each rock exerts a pushing force of 7.5 N.

Each rock exerts a pushing force of 15 N.
Physics
2 answers:
trapecia [35]3 years ago
7 0

Both would go 15N. Hoped I helped!

Marina CMI [18]3 years ago
4 0

Answer:

<h2>Each rock exerts a pushing force of 15 N.</h2>

Explanation:

The clue here is that the astronaut pushes with a force of 15N.

To understand this better we need to recur to Newton's third law which states

For every action (force), there's an equal and opposite reaction.

So, we know that the astronaut pushed both rocks with 15N force. By Netwon's third law, both rocks exert a force of 15N back, because such reaction must be equal.

Therefore, the right answer is the last choice: <em>Each rock exerts a pushing force of 15 N.</em>

<em />

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Determine the projection (magnitude and sign), or component, of vector v1 along the direction of vector v2. Your answer could be
professor190 [17]

Answer:

- 1.07 ft

Explanation:

V1 = (-5, 7, 2)

V2 = (3, 1, 2)

Projection of v1 along v2, we use the following formula

=\frac{\overrightarrow{V1}.\overrightarrow{V2}}{V2}

So, the dot product of V1 and V2 is = - 5 (3) + 7 (1) + 2 (2) = -15 + 7 + 4 = -4

The magnitude of vector V2 is given by

= \sqrt{3^{2}+1^{2}+2^{2}}=3.74

So, the projection of V1 along V2 = - 4 / 3.74 = - 1.07 ft

Thus, the projection of V1 along V2 is - 1.07 ft.

so we need to find the direction of v2

7 0
3 years ago
Most ocean waves obtain their energy and motion from _____. the moon's gravitational attraction the sun plate movement the wind
IrinaVladis [17]
Hello!

Most ocean waves obtain their energy and motion from the wind.

Ocean waves are surface waves that move across the surface of the ocean. When wind touches the surface of the water, there is friction in the contact zone. This friction causes a drag effect, that makes wrinkles on the surface of the water. As the wrinkles get bigger, they transform into full-blown waves, and the taller the wave, the more energy it can extract from the wind, making them even bigger and to move longer distances. 

Have a nice day!


3 0
3 years ago
During 57 seconds of use, 330 C of charge flow through a microwave oven. Compute the size of the electric current.
swat32

Answer:

5.78amps

Explanation:

Given data

Time t= 57 seconds

Charge Q= 330C

Current I= ??

The expression for the electric current is given as

Q= It

Substituting we have

330= I*57

I= 330/57

I=5.78 amps

Hence the current is 5.78amps

3 0
3 years ago
A race car has a maximum speed of 0.104 km/s .What is this speed in miles per hour ?
sweet [91]

Answer:

232.641374 mph

Explanation:

A race car has a maximum speed of 0.104km/s

Let X represent the speed in miles per hour

Therefore the speed in miles per hour can be calculated as follows

1 km/s = 2,236.936292 mph

0.104km/s = X

X = 0.104 × 2,236.936292

X = 232.641374

Hence the speed in miles per hour is 232.641374 mph

8 0
3 years ago
A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through a
Crank

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.

For the particular case of the Cube with equal sides its volume is determined by

V_c = l^3

V_c = 6^3 = 216cm^3

In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say

V_d = \pi r^2*l

V_d = \pi (\frac{2}{2})^2*6

V_d = 6\pi cm^3

In this way the net volume would be

\Delta V = V_c-V_d

\Delta V = 216cm^3-6\pi cm^3

\Delta V = 197.15cm^3 = 197.15*10^{-6}m^3

We need to find the mass, but we have the Weight and Gravity so from Newton's second Law

F= mg

m = \frac{F}{g}

m = \frac{6.6}{9.8}

m = 0.673kg

PART A) From the relation of density as a unit of mass and volume we have to

\rho = \frac{m}{V}

\rho = \frac{0.673}{197.15*10^{-6}}

\rho = 3413.64kg/m^3

PART B) To find the weight of the cube then we apply the ratio of

W = mg

W = V\rho g

W = (216*10^{-6})(3413.64)(9.8)

W = 7.22N

3 0
3 years ago
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