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r-ruslan [8.4K]

2 years ago

15

A tennis ball and a solid steel ball of the same diameter are dropped at the same time. Ignoring the air resistance effects, whi

ch ball has the greatest acceleration?

1 answer:

Paha777 [63]2 years ago

6 0

Yes I have the answer

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A moving roller coaster speeds up with constant acceleration for 2.3s until it reaches a velocity of 35m/s. During this time, th

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Hope this answer helps: option B

Explanation:

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En el MCU, la aceleración es: a) constante b) Varía en módulo, direccion, y sentido c) constante solo en el módulo d) constante

JulijaS [17]

If you put it into English in the comments I would be more then happy to help you! Thank you!

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2 years ago

A characteristic in an experiment, such as how much food, how much space, how hot or how cold, is termed a ____________. (spelli

KATRIN_1 [288]

I think the term is a variable?

6 0

2 years ago

Read 2 more answers

A long solenoid that has 1,130 turns uniformly distributed over a length of 0.430 m produces a magnetic field of magnitude 1.00

sweet-ann [11.9K]

Given Information:

Number of turns = N = 1130 turns

Length of solenoid = L = 0.430 m

Magnetic field = B = 1.0x10⁻⁴ T

Required Information:

Current = I = ?

Answer:

I = 0.0302 A

Explanation:

The current flowing in the solenoid winding can be found using

I = BL/μ₀N

Where μ₀ is the permeability of free space, N is the number of turns, B is the magnetic field and L is the length of solenoid

I = 1.0x10⁻⁴*0.430/4πx10⁻⁷ *1130

I = 0.0302 A

or

I = 30.28 mA

6 0

2 years ago

A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget

mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, $m_{1}$ = 4 kg

Speed, $v_{1}$ = 5 m/s

$m_{2}$ = 1 kg

$v_{2}$ = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

$\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2$

By plugging in the values we get,

$KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\$

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

$KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)$

$KE = 40J + KE(lost)$

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0

3 years ago