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Nitella [24]
3 years ago
5

A 20-kg wagon accelerates on a horizontal surface at 0.50 m⁄s2 when pulled by a rope exerting a 120-N force on the wagon at an a

ngle of 25° above the horizontal. Determine the magnitude of the effective friction force exerted on the wagon and the effective coefficient of friction associated with this force.

Physics
1 answer:
inn [45]3 years ago
5 0

Answer:

Explanation:

given,

mass of wagon = 20 Kg

horizontal acceleration = 0.5 m/s²

force of pull = 120 N

angle with horizontal = 25°

Applying newtons second law

F cos θ - f = m a

f = F cos θ- ma

f = 120 cos 25° - 20 x 0.5

f = 99 N

we know,

f = μ N

μ =\dfrac{f}{N}

μ =\dfrac{99}{mg - Fsin \theta}

μ =\dfrac{99}{20\times 9.8 - 120 sin 25^0}

μ = 0.68

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dangina [55]

Answer:

Explanation:

In physics, there are two types of physical quantities namely the fundamental and the derived quantities. Fundamental quantities are independent quantities on which derived quantities depends on. Length, mass and time are examples of fundamental quantities.

The SI unit of length is meters. A meter is a multiple unit. Its submultiple units are centimetres (10⁻²metres), kilometres (10³metres), decimetres (10⁻¹metres) etc

The SI unit of mass is kilogram (kg). The only sub multiple unit used in real-life situation is grams.

1 kg = 100 grams

The SI unit of time is seconds. The multiple units are the minutes, hours, weeks, days and years.

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4 0
3 years ago
At Alpha Centauri's surface, the gravitational force between Alpha Centauri and a 2 kg mass of hot gas has a magnitude of 618.2
sergeinik [125]

Answer:

6.86 * 10^8 m

Explanation:

Parameters given:

Mass of hot gas, m = 2 kg

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The gravitational force between two masses (the hot gas and Alpha Centauri) , m and M, at a distance, r, given as:

F = (G*M*m) / r²

Where G = gravitational constant

Therefore,

618.2 = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / r²

=> r² = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / 618.2

r² = 4.699 * 10^17 m²

=> r = 6.86 * 10^8 m

We are told that the hot gas is on the surface of Alpha Centauri, hence, the distance between both their centers is the radius of Alpha Centauri.

The mean radius of Alpha Centauri is 6.86 * 10^8 m.

5 0
3 years ago
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Answer:

<u>B) mass of pendulum bob</u>

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3 years ago
Consider projectile thrown horizontally at 50 m/s from height of 19.6 meters. The projectile will take ______________ time to hi
aleksley [76]

Answer:

C)The Same

Explanation:

Kinematics equation:

y=v_{oy}*t+1/2*g*t^2

for both cases the initial velocity in the axis Y is the same, equal a zero.

So the relation between the height ant temps is the same for both cases (the horizontal velocity does not play a role)

C)The Same

3 0
3 years ago
A 3.9 kg ball traveling towards a soccer player at a velocity of -3.5 m/s rebounds off the soccer player's foot at a velocity of
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Answer: 2.92 s

Explanation:

Given

Mass of ball is m=3.9\ kg

The initial velocity of the ball is u=-3.5\ m/s

Velocity after the rebound is v=15.9\ m/s

Force during the contact is F=25.9\ N

We know, change in momentum is Impulse

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3 years ago
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