The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration
The answer is below
Let g = acceleration due to gravity = 9.81 m/s², x = half of the width of the crate, half of the height of the crate = 0.5 m, a = acceleration of crate, N = force raising the crate
The sum of moment is given as:
Sum of vertical forces is zero, hence:
Sum of horizontal force is zero, hence:
Solving equation 1, 2 and 3 simultaneously gives :
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. If g-9.8 m/s, what is the coefficient of static friction? a. 3.0 b. 0.15 c. 0.28 d. 0.31
horizontal force, F = 750 N
mass of crate, m = 250 kg
g = 9.8 m/s^2
The friction force becomes applied force = 750 N
According to the laws of friction,
Friction force = μ x Normal reaction of the surface
here, μ be the coefficient of friction
750 = μ x m g
750 = μ x 250 x 9.8
μ = 0.31
Thus, the coefficient of static friction is 0.31.
Based on the calculatins, the coefficient of static friction is equal to: D. 0.31.
Given the following data:
Force = 750 Newton.
Mass = 250 kg.
Acceleration due to gravity = 9.8
How to calculate the coefficient of static friction.
Mathematically, the staticfrictional force acting on an object is giving by this formula:
is the coefficient of static friction.
N is the normal force.
m is the mass.
g is the acceleration due to gravity.
Substituting the given parameters into the formula, we have;
A 5-kg moving at 6 m/s collided with a 1-kg ball at rest. The ball bounce off each other and the second ball moves in the same direction as the first ball at 10 m/sec. What is the velocity of the first ball after the collision?
A 5-kg moving at 6 m/s collided with a 1-kg ball at rest.
The ball bounce off each other and the second ball moves in the same direction as the first ball at 10 m/sec.
To Find :
The velocity of the first ball after the collision.
We know, by conservation of momentum :
Putting all given values with directions ( one side +ve and other side -ve ).
Therefore, the velocity of first ball after the collision is 4 m/s after in opposite direction.
Hence, this is the required solution.
A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.20 107 m/s and experiences an acceleration of 1.90 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.
The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.
We need to calculate the magnetic field
Using formula of magnetic field
Using newton's second law
From equation (I) and (II)
Put the value into the formula
We need to calculate the direction of the field
Using the right hand rule, point the right hand fingers along the velocity which is in the positive z direction.
Now, if we curl the fingers along the direction of magnetic field that is in the negative y direction, then the thumb will point in the positive x direction.
Hence, The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.
Why are continental rocks much older than oceanic crust?A. Oceanic crust is continually recycled through convection in the earth's mantle B. Oceanic crust is made out of much less dense material than continental crust C. Continental crust is continually renewed through convection in the earth's mantle D. Continental crust eats oceanic crust for breakfast
A. Oceanic crust is continually recycled through convection in the earth's mantle
The oceanic plate is constantly being recycled through the forces of convection within the earth's mantle.
New oceanic plate are formed mid-oceanic ridge for example. As the magma cools and solidifies, they are moved away continually.
This is not the case for the continental curst.
Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.380 m. The engine of the car applies a torque of 456 N·m to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?
The magnitude of the static frictional force is 1200 N
given information :
radius, r = 0.380 m
applied-torque, τ1 = 456 N
The car has a constant velocity, thus the acceleration is zero
α = 0
Στ = I α
τ1 - τ2 = I α
τ2 = counter-torque
τ1 - τ2 = 0
τ1 = τ2
r x = τ1
= the static frictional force (N)
= τ1 /r
= 456 N/0.380 m
= 1200 N
An electromagnetic wave is traveling the +y direction. The maximum magnitude of the electric field associated with the wave is Em, and the maximum magnitude of the magnetic field associated with the wave is Bm. At one instant, the electric field has a magnitude of 0.25 Em and points in the +x direction. At this same instant, the wave's magnetic field has a magnitude which is ... and it is in the ... direction.
Direction is towards +z axis.
E = cB
E = magnitude of electrical 0.25 Em
c = speed of light in a vacuum 3x10^8 m/s
B = E/c = 0.25 ÷ 3x10^8
B = 0.83x10^-9 T
Magnetic fueld of a EM wave acts perpendicularly to its electric field, therefore it's direction is towards the +Z axis