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bulgar [2K]
3 years ago
13

where would a bowling ball and a napkin fall with the same acceleration. Check all that apply. A. Dropped off of the Leaning Tow

er of Pisa. B. on the moon. C. In a vacuum chamber. D. anywhere that is not very windy.
Physics
1 answer:
Kazeer [188]3 years ago
7 0

Answer: here ya go

A Bowling ball and a napkin would fall with the same acceleration in space where there is vacuum. Basically where there is no drag force and the two objects are freely falling under gravity. The two objects would fall with same acceleration due to gravity.

Explanation:

You might be interested in
Hermal energy is a form of... A. Potential energy. B. Kinetic energy. C. Chemical energy.
aev [14]

Thermal energy also known as 'heat energy' is a form of kinetic energy.

6 0
3 years ago
An object starts from rest at the origin and moves along the x axis with a constant
Alika [10]

Answer:

6.9m/s

Explanation:

Given parameters:

Acceleration of the object  = 4m/s²

Distance = from x; 2m to x; 8m

Unknown:

Average velocity  = ?

Solution:

From the given parameters, we use the right motion equation to solve the problem.

   v² = u² + 2aS

v is the final velocity

u is the initial velocity

a is the acceleration

 S is the distance covered

 Distance  = 8m - 2m  = 6m

 Initial velocity  = 0m/s

The final velocity gives us the average velocity in this problem;

       v² = 0² + (2 x 4 x 6) = 48

      v = √48  = 6.9m/s

7 0
3 years ago
A pickup truck has a width of 79.0 in. If it is traveling north at 42 m/s through a magnetic field with vertical component of
bekas [8.4K]

Answer:

The magnitude of the induced Emf is 0.003371V

Explanation:

The width of the truck is given as 79inch but we need to convert to meter for consistency, then

The width= 79inch × 0.0254=2.0066 metres.

Now we can calculate the induced Emf using expresion below;

Then the induced EMF= B L v

Where B= magnetic field component

L= width

V= velocity

=(40*10^-6) × (42) × (2.0066)

=0.003371V

Therefore, the magnitude emf that is induced between the driver and passenger sides of the truck is 0.003371V

8 0
4 years ago
Can you help me with this pls
devlian [24]

Answer:

Can you help me with this pls

4 0
3 years ago
Ple can someone help me to answer this question
ra1l [238]

The length of the uniform meter stick is 50 cm and the mass of the uniform meter stick is 50 g.

<h3 /><h3>A sketch of the uniform stick and the mass</h3>

The sketch of the uniform stick and the mass will be two based on the given statement.

<h3>when the uniform stick balances on knife at 10 cm from one end;</h3>

-------------------------------------------------------  

↓      10cm           Δ       (L - 10 cm)         ↓

200g                                                      M

<h3>When the knife edge is moved 5 cm further</h3>

|---------------15 cm-----------------|

-----------------------------------------------------------------------

              ↓      8.75 cm           Δ       (L - 15 cm)         ↓

          200g                                                               M

From the first diagram, apply principle of moment;

200(10) = M(L - 10) ------- (1)

From the second diagram, apply principle of moment;

200(8.75) = M(L - 15)   ------- (2)

From equation (1); M = (2000) / (L - 10)

From equation (2); M = (1750) / (L - 15)

Solve (1) and (2);

(2000) / (L - 10) = (1750) / (L - 15)

1750(L - 10) = 2000(L - 15)

L - 10 = 2000/1750(L - 15)

L - 10 = 1.143(L - 15)

L - 10 = 1.143L - 17.14

17.14 - 10 = 1.143L - L

7.14 = 0.143L

L = 7.14/0.143

L = 50 cm

<h3>Mass of the uniform stick</h3>

M = (2000) / (50 - 10)

M = 50 g

Thus, the length of the uniform meter stick is 50 cm and the mass of the uniform meter stick is 50 g.

Learn more about length of meter stick here: brainly.com/question/17225736

#SPJ1

8 0
2 years ago
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