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dangina [55]
3 years ago
13

Mation about your sour

Physics
1 answer:
Ann [662]3 years ago
8 0

(#1). (D).

(#2). (C).

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The coefficient of static friction between hard rubber and normal street pavement is about 0.90. On how steep a hill (maximum an
jok3333 [9.3K]
First thing to do is to draw the system described above. Then, write an equation for the forces present.
<span>
</span>Σ<span>F = Fg - Ff
</span><span>0 = mgsin</span><span>∅</span><span> - umgcos</span><span>∅</span><span>0 = gsin</span><span>∅</span><span> - ugcos</span><span>∅</span><span>
u = tan</span><span>∅
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7 0
3 years ago
Read 2 more answers
What is the volume of this bubble when it reaches the surface?
steposvetlana [31]

Answer:

Volume will be 15 mL. Solution:- If we look at the given information then it is Boyle's law as the temperature is constant and the volume changes inversely as the pressure changes. So, the volume of the air bubble at the surface will be 15 mL.

8 0
3 years ago
A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 4 kilograms is tied to the middle of the cloth
Olegator [25]

Answer:

T=26.03 N

Explanation:

Given that

Distance between poles = 12 m

Mass of block m= 4 kg

Sag distance = 5 m

Lets take tension in the clothesline is T.

The component of tension in vertical direction will be T cosθ.

By force balancing

2 T cosθ = 40

here tan\theta =\dfrac{5}{6}

θ=39.80°

2 T cos39.8 = 40

T=26.03 N

6 0
3 years ago
The masses of the two moons are determined to be 2M2M for Moon AA and MM for Moon BB . It is observed that the distance between
seraphim [82]

Answer:

 F_A = 8 F_B

Explanation:

The force exerted by the planet on each moon is given by the law of universal gravitation

        F = G \frac{m M}{r^{2} }

where M is the mass of the planet, m the mass of the moon and r the distance between its centers

let's apply this equation to our case

Moon A

the distance between the planet and the moon A is r and the mass of the moon is 2m

        F_A = G \frac{2m M}{r^{2} }

Moon B

        F_B = G \frac{m M}{(2r)^{2} }

         F_B = G \frac{m M}{4 r^{2} }

the relationship between these forces is

         F_B / F_A = \frac{1}{2 \ 4 } = 1/8

         F_A = 8 F_B

7 0
3 years ago
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Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 3.56 kg and rotate with
Natali5045456 [20]

Answer:

(a) 3107.98 J

(b) 14530.6 J

Explanation:

mass, m = 3.56 kg

angular speed, ω = 179 rad/s

Moment of inertia of solid cylinder, I = 1/2 mr^2

where, m is the mass and r be the radius of the cylinder.

(a) radius, r = 0.330 m

I = 0.5 x 3.56 x 0.330 x 0.330 = 0.194 kgm^2

The formula for the rotational kinetic energy is given by

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.194 x 179 x 179 = 3107.98 J

(b) radius, r = 0.714 m

I = 0.5 x 3.56 x 0.714 x 0.714 = 0.907 kgm^2

The formula for the rotational kinetic energy is given by

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.907 x 179 x 179 = 14530.6 J

3 0
3 years ago
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