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2 years ago

HEY EVERYONE!!! WATCH THIS!!!

Physics

2 answers:

VARVARA [1.3K]2 years ago

3 0

Answer:

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Explanation:

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Juliette [100K]2 years ago

3 0

Answer:

i swear to God, if I get Rick rolled...

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When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play

saw5 [17]

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

6 0

2 years ago

Suppose that you have been chosen for a space mission to a distant planet. Due to the length of time you'll be away from Earth y

Contact [7]

Answer:

I should be active for 15 hours to meet the physical activity requirement.

Explanation:

Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.

So, t = t₀/√(1 - β²)

t = 9/√(1 - (v/c)²)

= 9/√(1 - (0.8c/c)²)

= 9/√(1 - (0.8)²)

= 9/√(1 - (0.64)

= 9/√0.36

= 9/0.6

= 15 hr

So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.

8 0

2 years ago

Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

vivado [14]

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

where the relation between time (t) and period (T) is:

$\mathsf{t = \dfrac{T}{2}}$

T = 2t

T = 2(0.247)

T = 0.494 seconds

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

By making the spring constant k the subject of the formula:

$\mathsf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}$

$\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}$

$\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}$

$\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\ \mathsf{ k = \dfrac{2 \pi^2*m}{T^2}}$

$\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}$

$\mathbf{ k =8.25 \ N/m}$

Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

brainly.com/question/17315536?referrer=searchResults

6 0

2 years ago

an airplane is taking off from an airport with velocity of 200 km/h at an angle of 30 degrees above horizontal how fast is its r

ddd [48]

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7 0

2 years ago

You position two plane mirriors at right angles to each other a light ray strikes one mirrior at an angle of 60 to the normal an

Dafna1 [17]

Answer:

30 degrees

Explanation:

Reflects off of mirror 1 at 60 degrees....this makes it incident to second mirror at 30 degrees ....then angle of reflection equals this angle of incidence = 30 degrees

See atached diagram

4 0

1 year ago