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expeople1 [14]
3 years ago
5

During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel?

Physics
1 answer:
SIZIF [17.4K]3 years ago
4 0

-70560, -9.8/2 = -4.9

120^2 = 14400.

Now you multiply 14400 * -4.9 = -70560m.

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Convert 1nanosecond in to its SI init
SOVA2 [1]

<em>Convert 1nanosecond in to its SI init</em>

<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.</em>

<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.So to convert nanosecond into second, just multiply the nanosecond with 0.000000001 or (10^-9)</em>

8 0
2 years ago
The Burj Khalifa in Dubai is the world's tallest building. The structure is 828 m 828 m (2716.5 ft) and has more than 160 storie
Natasha_Volkova [10]

Answer:

h = height of the hotel room from the ground floor = 237.4m

Explanation:

Change in Potential Energy of tourist = ΔPE = PE2 – PE1 = mgh

PE1 is the potential energy of tourist at the ground floor

PE1 is the potential energy of tourist at the top (hotel room)

Given

PE1 = − 2.01 × 10⁵ J

PE2 = 0J

PE2 – PE1 = mgh

0 – (− 2.01 × 10⁵ J) = mgh

2.01 × 10⁵ J = 86.4×9.8×h

h = 2.01 × 10⁵/(86.4×9.8) = 237.4m

8 0
2 years ago
A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The
marusya05 [52]

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

3(670nm) /5 =402nm

∴λ = 402nm

Explanation:

4 0
3 years ago
What's the energy transfer for a catapult
just olya [345]
Kinetic energy is the energy for a catapult.
3 0
3 years ago
Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W
jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0
3 years ago
Read 2 more answers
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