Facebook
ClickBank1
ClickBank1

Volunteers/Contributors Needed!!


It is harder for me to update the site. I am looking for contributors who would want to post their physics answers on this site to help the community. I am sure I helped out a lot of people keeping things organized as they are. If any individuals would like to help out the site please leave a comment below. This is also a great study tool and database to use for your classes.

 

Best Wishes Everyone.

A 1-kg of wood is pressed against a vertical wall (µs= 0.25) by a force inclined 40º upward with respect to the vertical axis.

A 1-kg of wood is pressed against a vertical wall (µs= 0.25) by a force inclined 40º upward with respect to the vertical axis. Determine the maximum force which can start the motion of the block.

 

F =

(0.25)(9.8 cos40)+(9.8 sin30) / cos(4) – 0.25(sin30)=
7. 77 N

The coefficient of friction between the sole of a shoe and a tile span roof is 0.36.

The coefficient of friction between the sole of a shoe and a tile span roof is 0.36. What minimum slope of the roof can start slipping?

 

Θ = tan Θ = tan 20° =

0.36

A 60 kg package rests on the level floor of a warehouse.

1.) A 60 kg package rests on the level floor of a warehouse. If the coefficients of static and kinetic frictions are 0.56 and 0.37 respectively, what horizontal pushing force is required to
(a) start the package’s motion?
(b) slide the package across the floor at constant velocity?

 

60kg at rest
Find all the forces on the object
There is gravity…and Normal force pushing upward
Since the normal force opposes gravity and the object is not acclerating, the normal force = the force of gravity = mg

mg = 60kg x 9.8m/s^2 = 588N

mu = coefficient of friction
Ff=Force of friction
Fn = Force normal

mu = Ff/Fn

mu= 0.56
Fn= 588N

0.56 = Ff / 588N
0.56 x 588N = Ff
Ff = 329.28
The force of friction holding the object in place is 329.28N. NOTE that there is no force of friction YET because the object is at REST. But that’s how much resistance friction can apply
SO you need a force GREATER THAN 329.28N to START its motion

mu = Ff/Fn
0.37 = Ff/588N
0.37 x 588N = Ff
Ff = 217.56N
The maximum force of friction that can be applied at motion is 217.56N
To keep at a constant velocity, the force has to = 217.56

A 1340 kg Triumph TR-8 traveling north at 43 km/hr crashes head-on into a 1850 kg Checker cab moving west at 87 km/hr.

A 1340 kg Triumph TR-8 traveling north at 43 km/hr crashes head-on into a 1850 kg Checker cab moving west at 87 km/hr. The two cars become tangled together. What is their speed a moment after the collision?

If east is 0 degrees, north is 90 degrees, etc., at what angle are the entangled cars travelling just after their collision?

 

Let
m1 = 1340 Kg; v1 = 43 km/h; m2 = 1850 kg; v2 = 87 km/h
mpmentum of m1 = P1; momentum of m2 = P2

I’ll break up the momenta into their vertical and horizontal components. I’ll use the sign convention that up and left are positive, down and right are negative.

P1 has no horizontal component so…
P1 = (m1v12) ĵ

P2 has no vertical component so…
P2 = (m2v22) î

so

P = (m2v22) î + (m1v12) ĵ

From Pythagoras

I P I = √ [(m2v12)2 + (m1v12)2]
I P I = 14220162 Kg m/s

This is equal to the mass * velocity2 of the cars after collision.

v^2 = 14220162 / (1850 + 1340)
v = √[14220162 / 3190]
v = 66.7 m/s

The direction of the momentum is given by

arc tan [(m1v12) / (m2v22)] = arc tan (0.177) = 10 degrees north of west.

If east is 0 degrees then the angle is 180 – 10 = 170 degrees

A charge of -7 x 10-8 C exerts a force of 0.045 N on a charge of +6.0 x 10-7.

A charge of -7 x 10-8 C exerts a force of 0.045 N on a charge of +6.0 x 10-7.

How far apart far are the charges?

 

f=kQq/r2
Rearanged: [sqrt](kQq/r2)

=[√](9×109)(7×10-8C)(6×10-7C)/(0.…
r=.092m

In reaching her destination, a backpacker walks with an average velocity of 1.27 m/s, due west.

In reaching her destination, a backpacker walks with an average velocity of 1.27 m/s, due west. This average velocity results, because she hikes for 5.89 km with an average velocity of 2.18 m/s due west, turns around, and hikes with an average velocity of 0.555 m/s due east. How far east did she walk (in kilometers)?

You need to find the distance east, you know the velocity east so you just need to find the time spent traveling east, using the average velocity formula

take west as positive direction, east as negative

V = 1.27m/s
v1 = 2.18m/s
t1 = 5890/2.18 = 2702s …..time traveling west
v2 = -0.555m/s
t2 = ?………………………….time traveling east

d1 = v1t1 = 5890m ……….displacement west
d2 = v2t2 = -0.555t2………displacement east

final displacement is
D = d1 + d2 = 5890 – 0.555t2
the total time of travel is
T = t1 + t2 = 2702 + t2
the average velocity is
V = 1.27 = D/T= (5890 – 0.555t2)/ (2702 + t2)
gives
1.27(2702 + t2) = (5890 – 0.555t2)
expand and rearrange for t2
t2 = (5890 – 1.27*2702) / (1.27 + 0.555) = 1347s
is the time spent traveling east

so the distance traveled east is
|d2| = |v2t2| = 0.555*1347 = 748m = 0.748km

The froghopper, Philaenus spumarius, holds the world record for insect jumps.

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0 degrees above the horizontal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground. The initial velocity is 4 m/s. What horizontal distance did the froghopper cover for this world-record leap?

 

Use the range equation:
range = sin2(θ)v2/9.8 where v is launch velocity

and
theta is the launch angle
range = sin2(58)16/9.8
range = 1.624 m

A man is laying on a board elevated by a support at the tip of the head of the board and by a scale at the foot of the board.

A man is laying on a board elevated by a support at the tip of the head of the board and by a scale at the foot of the board. 
What would happen if a support is placed exactly at x = 79 cm (which is the center of gravity) followed by the removal of the supports at the subject’s head and feet? (Assume the board is removed along with the supports. Select all that apply.)

A)The subject’s center of gravity would shift toward his head. 
B)The subject’s head would sink toward the ground. 
C)The subject’s feet would sink toward the ground. 
D)The subject’s center of gravity would not shift. 
E)The subject’s center of gravity would shift toward his feet. 
F)The subject would remain balanced.

 

Since you are removing the board, the subject is not ridged. Thus, his body will flex roughly into a curve over the balance point. If there was enough room for that curve he would remain balanced >

B, C, D, F are all still true

A hunter aims directly at a target (on the same level) 140 m away. If the bullet leaves the gun at a speed of 280 m/s, by how much will the bullet miss the target?

A hunter aims directly at a target (on the same level) 140 m away. If the bullet leaves the gun at a speed of 280 m/s, by how much will the bullet miss the target?

 

In projectile motion constant horizontal velocity Vx is assumed.

Time to reach target is T = 140m / Vx = 0.5 s

Vertical deflection in time T = 0.5*g*T2 = 0.5*9.8*0.52 m = 1.23 m

So the bullet passes 1.23 m under the target.