The post We Are Back appeared first on Physics Mastered - Master Physics.

]]>As you can see now, the site sas a new format similar to Yahoo answers. If you have a problem that you are stuck on, ask away and hopefully, someone with the solution can give you a hand.

Tell your friends about the site and let’s build a community where people can come in and ask for help and give back at the same time.

The post We Are Back appeared first on Physics Mastered - Master Physics.

]]>The post Volunteers/Contributors Needed!! appeared first on Physics Mastered - Master Physics.

]]>

Best Wishes Everyone.

The post Volunteers/Contributors Needed!! appeared first on Physics Mastered - Master Physics.

]]>The post 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. appeared first on Physics Mastered - Master Physics.

]]>

F =

(0.25)(9.8 cos40)+(9.8 sin30) / cos(4) – 0.25(sin30)=

7. 77 N

The post 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. appeared first on Physics Mastered - Master Physics.

]]>The post The coefficient of friction between the sole of a shoe and a tile span roof is 0.36. appeared first on Physics Mastered - Master Physics.

]]>

Θ = tan Θ = tan 20° =

0.36

The post The coefficient of friction between the sole of a shoe and a tile span roof is 0.36. appeared first on Physics Mastered - Master Physics.

]]>The post A 60 kg package rests on the level floor of a warehouse. appeared first on Physics Mastered - Master Physics.

]]>

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

The post A 60 kg package rests on the level floor of a warehouse. appeared first on Physics Mastered - Master Physics.

]]>The post 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. appeared first on Physics Mastered - Master Physics.

]]>**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 = (m1v1^{2}) ĵ

P2 has no vertical component so…

P2 = (m2v2^{2}) î

so

P = (m2v2^{2}) î + (m1v1^{2}) ĵ

From Pythagoras

I P I = √ [(m2v1^{2})^{2} + (m1v1^{2})^{2}]

I P I = 14220162 Kg m/s

This is equal to the mass * velocity^{2} 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 [(m1v1^{2}) / (m2v2^{2})] = arc tan (0.177) = 10 degrees north of west.

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

The post 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. appeared first on Physics Mastered - Master Physics.

]]>The post A charge of -7 x 10-8 C exerts a force of 0.045 N on a charge of +6.0 x 10-7. appeared first on Physics Mastered - Master Physics.

]]>**How far apart far are the charges?**

f=kQq/r^{2}

Rearanged: [sqrt](kQq/r^{2})

=[√](9×10^{9})(7×10^{-8}C)(6×10^{-7}C)/(0.…

r=.092m

The post A charge of -7 x 10-8 C exerts a force of 0.045 N on a charge of +6.0 x 10-7. appeared first on Physics Mastered - Master Physics.

]]>The post In reaching her destination, a backpacker walks with an average velocity of 1.27 m/s, due west. appeared first on Physics Mastered - Master Physics.

]]>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 post In reaching her destination, a backpacker walks with an average velocity of 1.27 m/s, due west. appeared first on Physics Mastered - Master Physics.

]]>The post The froghopper, Philaenus spumarius, holds the world record for insect jumps. appeared first on Physics Mastered - Master Physics.

]]>

Use the range equation:

range = sin2(**θ**)v^{2}/9.8 where v is launch velocity

and

theta is the launch angle

range = sin2(58)16/9.8

range = 1.624 m

The post The froghopper, Philaenus spumarius, holds the world record for insect jumps. appeared first on Physics Mastered - Master Physics.

]]>The post 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. appeared first on Physics Mastered - Master Physics.

]]>**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

The post 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. appeared first on Physics Mastered - Master Physics.

]]>