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Dovator [93]
3 years ago
15

You are asked to design an experiment which is similar to Match Graph activity. If you move 3m back from the motion sensor in 2s

. What is your speed?
Physics
1 answer:
Art [367]3 years ago
5 0

Answer:

The speed is 1.5 m/s.

Explanation:

Given that,

Distance = 3 m

Time = 2 sec

We know that,

The position and time graph shows the speed of motion.

We need to calculate the speed

Using formula of speed

v=\dfrac{d}{t}

Where, d = distance

t = time

Put the value into the formula

v=\dfrac{3}{2}

v=1.5\ m/s

Hence, The speed is 1.5 m/s.

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A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4.8 m/s
yKpoI14uk [10]
  • initial velocity=0m/s=u
  • Acceleration=a=4.8m/s^2
  • Time=t=15s

Final velocity be v

\\ \sf\longmapsto v=u+at

\\ \sf\longmapsto v=0+4.8(15)

\\ \sf\longmapsto v=72m/s

8 0
2 years ago
How is a scientific law different than a scientific theory ?
OLga [1]

Answer:  scientific law is the description of an observed phenomenon. It doesn't explain why the phenomenon exists or what causes it. The explanation of a phenomenon is called a scientific theory

hope this helps

plz mark brainleist

3 0
3 years ago
the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is
cluponka [151]

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

3 0
3 years ago
Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side
pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m  + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0
3 years ago
How is frequency related to the sound we hear?
Leviafan [203]
Frequency is the vibration of noise and the vibration determines the pitch, which we depend on to be a pitch or frequency we can hear. If it's too high or too low our ears can't hear it 
8 0
3 years ago
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