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nadezda [96]
3 years ago
12

You are pulling a 80 kg box with a rope. The force you exert on the box is 90 N at 30 degrees from the ground. What would be the

coefficient of kinetic friction if the box is moving at constant velocity?
Physics
1 answer:
Feliz [49]3 years ago
7 0

Answer:

Coefficient of dynamic friction= md= 0.09931

Explanation:

To determine the coefficient of dynamic friction we must first match the friction force that is permendicular to the normal force of the block and opposite to the drag force, to the component of the drag force in this same direction. This component on the X axis of the drag force will be:

F= 90N × cos(30°) = 77.9423N

This component on the X axis of the drag force must be equal to the dynamic friction force that is equal to the coefficient of dynamic friction by the normal force of the block weight:

F= md × m × g= 77.9423N

m= mass of the block

md= coefficient of dynamic friction

g= gravity acceleration

F= md × 80kg× 9.81 (m/s²)= 77.9423(kg×m/s²)

md= (77.9423(kg×m/s²) / 784.8 (kg×m/s²)) = 0.09931

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James Bond is trying to escape his enemy on a speedboat but
SVEN [57.7K]

Answer:

100 m/s

Explanation:

Mass the mass of Bond's boat is m₁. His enemy's boat is twice the mass of Bond's i.e. m₂ = 2 m₁

Initial speed of Bond's boat is 0 as it won't start and remains stationary in the water. The initial speed of enemy's boat is 50 m/s. After the collision, enemy boat is  completely stationary. Let v₁ is speed of bond's boat.

It is the concept of the conservation of momentum. It remains conserved. So,

m_1u_1+m_2u_2=m_1v_1+m_2v_2

Putting all the values, we get :

0+(2m_1)50=m_1v_1+(2m_2)(0)\\\\100m_1=m_1v_1\\\\v_1=100\ m/s

So, Bond's boat is moving with a speed of 100 m/s after the collision.

3 0
3 years ago
What will happen if our atmosphere was oxygen
qaws [65]
Most humans would die do to the excess oxyegen
3 0
2 years ago
A wire carrying a current of 26.9 A is bent into a circular arc with a radius of 0.6 cm that sweeps out 0.900 radians. What is t
melomori [17]

The magnetic field at the center of the arc is 4 × 10^(-4) T.

To find the answer, we need to know about the magnetic field due to a circular arc.

<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>
  • According to Biot savert's law, magnetic field at the center of a circular arc is
  • B=(μ₀ I/4π)× (arc/radius²)
  • As arc is given as angle × radius, so

        B=( μ₀I/4π)×(angle/radius)

<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>

B=(μ₀ I/4π)× (0.9/0.006)

  = (10^(-7)× 26.9)× (0.9/0.006)

  = 4 × 10^(-4) T

Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.

Learn more about the magnetic field of a circular arc here:

brainly.com/question/15259752

#SPJ4

5 0
2 years ago
What simple machines are used to create the shovel?
erastovalidia [21]

Answer:

I believe a wedge and a lever

Explanation:

7 0
3 years ago
The current through a 10-ohm resistor connected to a 120-V power supply is The current through a 10-ohm resistor connected to a
Mazyrski [523]

Answer:

12 A

Explanation:

V= I * R

V/R = I

120 V /10 Ω  = 12 AMPS

7 0
1 year ago
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