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KATRIN_1 [288]
2 years ago
13

An iron block of mass 10 kg rests on a wooden plane inclined at 30° to the horizontal. It is found

Physics
1 answer:
Kaylis [27]2 years ago
5 0

I assume the 100 N force is a pulling force directed up the incline.

The net forces on the block acting parallel and perpendicular to the incline are

∑ F[para] = 100 N - F[friction] = 0

∑ F[perp] = F[normal] - mg cos(30°) = 0

The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.

Then

F[friction] = 100 N

F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N

If µ is the coefficient of static friction, then

F[friction] = µ F[normal]

⇒   µ = (100 N) / (84.9 N) ≈ 1.2

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Ymorist [56]

Answer:

The final velocity of the ball is 7m/s

Explanation:

M1=8kg,  V1 =10m/s , M2=2kg , V2=-5m/s

initial momentum before collison

m1v1+m2v2

=8×10 +2×(-5)  =80-10  = 70kg m/s

final momentum after collison

=(m1+m2)×v

=(8+2)×v

=10v

According to the law of conversion of momentum

initial momentum =final momentum

70=10v

10v=70

v=70/10

v=7m/s

3 0
3 years ago
plz i need help fast................a motion along a straight line with a constant increase in velocity is ?​
Monica [59]

Answer:

uniform acceleration

Explanation:

The definition for uniform acceleration is:

if an object travels in  a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration is said to be uniform.

Hope this helps.

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8 0
3 years ago
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How many poles do you expect to see in your magnet? Look around your room. Make a list of 5 items that might be attracted to the
jeyben [28]

Answer:

1.1 Two poles: North and South Poles.

1.2 - Staple pin - Nail - Tip of my phone charger - Metal keys - Cloth Hanger

1.3 - Wooden bed cot - Plastic pen - Game pad - Wooden shelf - Paper - A T-shirt

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3 years ago
A bullet B of mass mB traveling with a speed v0 = 1400 m/s ricochets off a fixed steel plate A of mass mA. Let mA ≫ mB so that i
Greeley [361]

Answer:

Rebounce angle is 345°

Rebounce speed is 989.95m/s

Explanation:

Calculate the x  component of the velocity of the bullet before impact by using the following relation:

Vbx= Vb Cos thetha

Here,  is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°

Substituting

Vbx = Cos15 ×1400 = 1352.30m/s

Calculate the y component using the relation:

Vby = Vo Sin theta

Vby = sin 15° × 1400

Vby = 362.35m/s

The rebounce angle = 360 - incidence angle

Rebounce angle =( 360 - 15)° = 345°

The rebound speed V' = Vby - Vbx

V' = (1352.30 - 362.35)m/s

V' = 989.95 m/s

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3 years ago
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