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scoundrel [369]
2 years ago
11

Which is an effect of gravity on objects on the surface of Earth? Check all that apply.

Physics
1 answer:
timama [110]2 years ago
3 0

• Objects “fall” toward the center of Earth.  (TRUE)

Since the gravitational force between earth and any object is always along the line joining the centers of two so all the objects near the surface of earth will always feel force towards the centre of the earth due to gravity.

• Objects are pulled by Earth but do not pull on Earth.  (FALSE)

Since gravitational force will always follow Newton's III law as per which every action has equal and opposite reaction force so object near the surface of earth will also pull the earth as earth is pulling the object towards the center.

• Objects accelerate at a rate of 9.8 m/s each second.  (TRUE)

Since acceleration due to gravity of Earth is 9.8 m/s^2 so here we can say that all objects near the surface of earth will accelerate towards the center of earth at rate of 9.8 m/s^2 due to gravitational pull.

• Objects travel at a rate of 9.8 m/s to are the center of Earth. (FALSE)

Since earth is pulling all objects near its surface with gravitational force so objects can not move at constant speed but it will move with constant acceleration.

• Objects are pulled by Earth more strongly than they pull on Earth. (FALSE)

Since gravitational force will always follow Newton's III law which say every action has equal and opposite reaction so we can say that force due to all objects on earth will be same as the force of earth on the object.

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A skateboarder traveling with an initial velocity 9.0 meters per second,
meriva

Answer:

25m/s

Steps:

<em> First, The equation v= u + a * t shows us what we need to find, (the finale velocity). </em>

<em />

Second, we substitute the values given:

v= 9m/s + 4m/s2 * 4s

Last, We calculate the values:

Multiply 4m/s2 * 4s = 16m/s  

Add 9m/s + 16m/s

<u></u>

<u>Answer:  25m/s</u>

Hope this helps :)

4 0
3 years ago
Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts
Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground  = 3.13 m/s

4 0
3 years ago
A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz
gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle?  Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa  

Tt is stagnation temperature = ?

T is the static temperature  = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa =  ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 =  Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0
3 years ago
Why is streak generally a more useful property than color in identifying a mineral?
poizon [28]
You may confuse two minerals by their colors so using streak is another way for showing their "true colors "
3 0
3 years ago
What is glacier flood?<br>​
posledela

Answer:

A glacial lake outburst flood is a type of outburst flood caused by the failure of a dam containing a glacial lake. An event similar to a GLOF, where a body of water contained by a glacier melts or overflows the glacier, is called a jökulhlaup. The dam can consist of glacier ice or a terminal moraine.

3 0
2 years ago
Read 2 more answers
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