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Bad White [126]
2 years ago
11

A battery-powered lawn mower has a mass of 48.0 kg. If the net external force on

Physics
1 answer:
SpyIntel [72]2 years ago
5 0
F= mass*acceleration
_40.5=48.0*a
a=40.5/48.0
a=0.84m/s
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If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
2 years ago
What's a streak plate?
andrew-mc [135]
An unglazed piece of porcelain, used to test the characteristic streak of minerals by rubbing the mineral across the tile. Streak plates have a hardness of about 6.5 on the Mohs scale and cannot be used for testing harder minerals.
3 0
3 years ago
PLEASE HELP ITS DUE TODAY ILL POST THE OTHER HALF
IgorLugansk [536]
2.B
4.C
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Theses are the right answer
4 0
2 years ago
Read 2 more answers
in a falling elevator if you jump at the right time would you be safe? explain, and give a reasonable response.
N76 [4]

Answer:

No

Explanation:

You can easily injury yourself due to the force of gravity.

6 0
3 years ago
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A satellite is put in a circular orbit about Earth with a radius equal to 35% of the radius of the Moon's orbit. What is its per
Debora [2.8K]

Answer:

0.21 lunar month

Explanation:

the radius of moon = r₁

time period of the moon = T₁ = 1 lunar month

The radius of the satellite = 0.35 r₁

Time period of satellite

The relation between time period and radius

              T\ \alpha\ \sqrt{r^3}

now,

              \dfrac{T_2}{T_1}=\dfrac{\sqrt{r_2^3}}{\sqrt{r_1^3}}

              \dfrac{T_2}{T_1}=\dfrac{\sqrt{0.35^3r_1^3}}{\sqrt{r_1^3}}

              \dfrac{T_2}{1}=\sqrt{0.35^3}

                              T₂ = 0.21 lunar month

hence, the time period of revolution of satellite is equal to 0.21 lunar month

6 0
3 years ago
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