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tia_tia [17]
3 years ago
8

Is this statement true or false? The pull of Earth’s gravity doesn’t affect the Moon.

Physics
2 answers:
Marta_Voda [28]3 years ago
7 0
False false false. Why else would to moon REVOLVE around the earth. It's all because of Earth's gravity
In-s [12.5K]3 years ago
6 0
That statement is false.  The gravitational attraction between
the Earth and the Moon is what keeps the Moon in Earth-orbit.
It's also the force that keeps the same side of the Moon always
facing the Earth.
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State ohms law and explain the terms associated with it​
stich3 [128]

Explanation:

Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit. To students of electronics, Ohm's Law (E = IR)

3 0
3 years ago
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A student forms the following hypothesis if people wear bright colors, they will be happier. What is/are the problems with this
mars1129 [50]
There is no scientific evidence to support this claim. And it is not telling what colors are needing to be worn only the stated bright colors which can mean an assortment of things.
8 0
3 years ago
Frosty the Snowman accelerates from 16 m/s to 26 m/s in 2 s. What is his acceleration?
Alex_Xolod [135]

Answer:

his acceleration is 3.25

Explanation:

because you do 26÷16 which =1.625×2=3.25

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3 years ago
Identical objects, Object X and Object Y, are tied together by a string and placed at rest on an incline, as shown in the figure
goblinko [34]

The slope of the velocity time graph of an object moving with constant acceleration is constant

It will take approximately <u>3 seconds</u> for the center of mass of Object X to reach point J near the bottom of the incline

The reason why the above time value is correct is given as follows:

Known parameters:

Initial velocity of the objects, u = 0

The graph  in the question is a straight line graph with data points

(0, 0), (0.5, 1.0), (1.0, 2), (3.0, 6), and (3.5, 7)

Given that the slope of the velocity-time graph is constant, we have that the acceleration is constant and is given as follows;

a = \dfrac{\Delta v}{\Delta t } = \dfrac{v_2 - v_1}{t_2 - t_1}

Therefore;

a =  \dfrac{6 - 2}{3.0 - 1.0} = 2

The acceleration, a ≈ 2 m/s²

The distance from the center of mass of the Object X to the point J near the bottom = 9 m

The equation for distance travelled is given as follows;

s = u\cdot t + \dfrac{1}{2} \cdot a \cdot t^2

Which gives;

9 =  0\times t + \dfrac{1}{2} \times 2 \times t^2 = t^2

t = √9 = 3

The time it will take the center of mass of Object X to reach point J near the bottom of the incline is t = <u>3 seconds</u>

Learn more about motion under constant acceleration here:

brainly.com/question/16391598

6 0
3 years ago
The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and
ira [324]

Complete question is;

a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?

b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be

Answer:

A) V_t = 18 m/s

B) V_t = 10.39 m/s

Explanation:

Formula for terminal speed is given by;

V_t = √(2mg/(DρA))

Where;

m is mass

g is acceleration due to gravity

D is drag coefficient

ρ is density

A is Area of object

A) Now, for sphere 1,we have;

m = 1 kg

V_t = 6 m/s

g = 9.81 m/s²

Now, making D the subject, we have;

D = 2mg/((V_t)²ρA))

D = (2 × 1 × 9.81)/(6² × ρA)

D = 0.545/(ρA)

For sphere 2, we have mass = 9 kg

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]

V_t = 18 m/s

B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.

Thus;

Area of sphere 2 = 3A

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]

V_t = 10.39 m/s

5 0
4 years ago
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