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Alexxx [7]
3 years ago
14

PLEASE HELP ASAP!!!!!!!!!!!!

Physics
1 answer:
AleksandrR [38]3 years ago
3 0

a = acceleration of swati during free fall = - 9.8 meter/second²

v₀ = initial velocity of swati = 0 meter/second

t = time of travel = 120 seconds

d = displacement of swati during free fall

Using the kinematics equation

d = v₀ t + (0.5) a t²

inserting the values

d = (0) (120) + (0.5) (- 9.8) (120)²

d = - 70560 m

hence distance travelled by swati is 70560 m

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A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
Marrrta [24]

Answer:

t = 6.09 seconds

Explanation:

Given that,

Speed, v = 44.1 cm/s

Distance, d = 269 cm

We need to find the time interval of the marble. Speed is distance per unit time.

v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s

Hence, the time interval of the marble is 6.09 seconds.

6 0
4 years ago
The mass of an object is 10 kg and the velocity is 4 m/s, what is the momentum?
eduard

The answer is 40 kg. m/s.

Formula for momentum:

p=mv

p=(10 kg.)(4 m/s)

So, therefore, the final answer is p=40 kg. m/s.

I hope this helped answer your question. Enjoy your day, and take care!

3 0
3 years ago
What is the wavelength in nm of a light whose first order bright band forms a diffraction angle of 30 degrees, and the diffracti
Artist 52 [7]

Answer:

The wavelength is 3500 nm.

Explanation:

d= \frac{1}{700 lines per mm} = 0.007mm = 7000 nm

n= 1

θ= 30°

λ= unknown

Solution:

d sinθ = nλ

λ = \frac{7000 nm sin 30}{1}

λ = 3500 nm

8 0
4 years ago
A 1.0 kg ball falling vertically hits a floor with a velocity of 3.0 m/s and bounces vertically up with a velocity of 2.0 m/s .
ale4655 [162]

Answer:

50 N

Explanation:

given,

mass of ball = 1 Kg

initial velocity = 3 m/s

final velocity = 2 m/s

time = 0.10 s

impulse = change in momentum

I = m ( v - u)            

I = 1( 2 -(-3))                

I = 5 kg.m/s          

Force is equal to impulse per unit time

average force = \dfrac{I}{t}

                       = \dfrac{5}{0.1}

                       = 50 N

Average force on the floor will be equal to 50 N

5 0
3 years ago
A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal
Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, a=g=-9.8\ m/s^2(Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

S=ut+\frac{1}{2}at^2

Plug in the given values and solve for 'u'. This gives,

0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0
4 years ago
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