Answer:
Before providing an answer to the question, the values for acceleration given in questions A and B were written twice. So correction would go like this: For (a) when the truck accelerates at 2.20m/s2 northward, and for (b) when it accelerates at 3.40m/s2 southward.
The answer:
(a) 88N, northward.
(b) 78.4, southward.
Explanation:
(a) Maximum frictional force acting on the packing case= (coefficient of static friction) X (Normal force)
Normal force = mass X acceleration due to gravity
Maximum static frictional force acting on the packing case = (coefficient of static friction) X (mass of packing case X acceleration due to gravity)
Maximum static frictional force = (0.30) X (40.0-kg) X (9.8m/s 2) = 117.6N
While Reaction force acting on the packing case = (mass of packing case) x (acceleration generated by the pickup truck)
Reaction force acting on the case = (40.0-kg) X (2.20m/s2) = 88N
With these values, one can conclude that the packing case is at rest since the reaction force of the case acting in the opposite direction is lesser than the frictional force. Making the magnitude and direction of the friction force acting on the case still move northward, and the static frictional force acting on equals the reaction force.
The answer is 88N, northward.
(b) Here too, we need to still compare the reaction force with the value of the already determined Maximum static frictional force (117.6N) above. This is necessary to know the frictional force between the pickup truck"s floor and the packing case.
Reaction force acting on the case when acceleration is 3.40m/s2 = (40.0-kg) X (3.40m/s2) = 136 N
We can conclude that the reaction force (136 N) is greater than the maximum static frictional force (117.6N), suggesting that the packing case is in motion and the frictional force is no longer static.
This means a kinetic force is now acting on the pickup truck"s floor causing the packing case to also move. This kinetic force can be calculated as:
kinetic force = (coefficient of kinetic friction) X (mass of packing case X acceleration due to gravity)
= (0,20) X (40.0-kg) X (9.8m/s 2) = 78.4N
HELP ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!!
The Left and right forces acting on this cart are
A. balanced
B. super-positioned
C. unbalanced
D. counter-balanced
Answer:
B. Super-Positioned
Explanation:
A small mailbag is released from a helicopter that is descending steadily at 2.00 m/s. After 3.00 s.
Required:
a. What is the speed of the mailbag?
b. How far is it below the helicopter?
Answer:
Explanation:
Initial velocity of mailbag u = 2 m/s
acceleration downwards a = g = 9.8 m/s²
time t = 3 s
a ) final velocity v = ?
v = u + at
= 2 + 9.8 x 3
= 31.4 m /s
b )
s = ut + 1/2 g t²
s is relative displacement of mailbag
u = relative initial velocity of mailbag = 0
relative acceleration = g = 9.8 m /s²
time t = 3 s
s = 0 + 1/2 x 9.8 x 3²
= 44.1 m
relative displacement of mailbag = 44.1 m .
If electrical current is moving through a horizontal wire toward your face, what direction is the induced magnetic field?
a. clockwise
b. straight left
c. straight right
d. counter clockwise
Answer:(b)
Explanation:
Given
Electric current flowing through the horizontal wire towards the observer's face.
The direction of the magnetic field is given by the right-hand thumb rule, i.e. place the thumb in the direction of current and the wrapping of fingers will give the direction of the magnetic field
the direction of the magnetic field will be counterclockwise as observed by an observer.
a car with a mass of 2050 kg is traveling at 4.5 m/s. what is the cars momentum
Answer:
9225 kg m/s
Explanation:
momentum = mass * velocity
momentum = 2050 * 4.5
momentum = 9225
If a force acts constantly on a stationary object, the object will
Answer:
accelerate in the direction of the force
Explanation:
You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, where the lumps collide and stick together. Just before the impact, the velocity of one lump was < 3, 3, -3 > m/s, and the velocity of the other lump was < -4, 0, -4 > m/s. What is the velocity of the stuck-together lump just after the collision
Answer:
[tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]
Explanation:
[tex]u_1[/tex] = Velocity of one lump = [tex]3x+3y-3z[/tex]
[tex]u_2[/tex] = Velocity of the other lump = [tex]-4x+0y-4z[/tex]
m = Mass of each lump = [tex]30\ \text{g}[/tex]
The collision is perfectly inelastic as the lumps stick to each other so we have the relation
[tex]mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]
The velocity of the stuck-together lump just after the collision is [tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex].
