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Nookie1986 [14]
3 years ago
7

A motor has a rating power of 20 W and motor is used for 10 s. Calculate the

Physics
1 answer:
zepelin [54]3 years ago
6 0

Explanation:

oii ay school ens?

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Chris threw a basketball a distance of 27.5 m to score and win his
salantis [7]

Answer:

v₀ = 16.55 m/s

Explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

θ = Launch Angle = 50°

g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}

<u>v₀ = 16.55 m/s</u>

8 0
3 years ago
The wetter the surface, the friction
allochka39001 [22]
Answer: Water can either increase or decrease the friction between surfaces.
7 0
3 years ago
A car enters a horizontal, curved roadbed of radius 50 m. the coefficient of static friction between the tires and the roadbed i
UkoKoshka [18]
Previous results tell us the speed (v) is given in terms of the coefficient of friction (k) and the radius of the curve (r) as
  v = √(kgr)
  v = √(0.20·9.8 m/s²·50 m)
  = 7√2 m/s ≈ 9.90 m/s
6 0
3 years ago
Read 2 more answers
A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su
madreJ [45]

maximum static friction acting on the object will be

F_s = \mu_s mg

plug in all values

F_s = 0.40 \times 15 \times 9.8 = 58.8 N

So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force

So here it is given that applied force is 20 N

so here object will not move due to this force and it will remain at rest always

due to this applied force

6 0
3 years ago
Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

8 0
3 years ago
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