let Coefficients of Friction of Rubber on asphalt (dry) =0.7
F= Coefficients of Friction * normal force = 0.7 * 60 =42 N
so the net force of the rubber is zero, meaning it will travel at a constant speed.
When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed
Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
Answer:
Usually the coefficient of friction remains unchanged
Explanation:
The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.
Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.
Answer:
) pulls the ladder in the direction opposite
Explanation:
This is in line with lenz law that states that the magnetic field induced in a conductor act to oppose the magnetic field that produced it
From rest, a rock is dropped from a garage roof. The roof is 6.0 meters above ground level. The rock will reach the earth at a speed of 10.849 meters per second.
<h3>What is velocity?</h3>
The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity.
it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.
Given data:
V(Final velocity)=? (m/sec)
h(height)= 6.0 m
u(Initial velocity)=0 m/sec
g(gravitational acceleration)=9.81 m/s²
Newton's third equation of motion:
Hence, the velocity of the rock as it hits the ground will be 10.849 m/sec.
To learn more about the velocity refer to the link ;
brainly.com/question/862972
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