PHYSICS COLLEGE

Name at 2 areas of physics that make video games possible

Answers

Answer 1

Answer:

Answer:

projectiles

electromagnetic

Answer 2

Answer:

Answer:

Explanation:

física cuántica y Quantum Moves

Related Questions

A block slides from rest with negligible friction down the track above, descending a vertical height of 5.0 m to point P at the bottom. It then slides on the horizontal surface. The coefficient of friction between the block and the horizontal surface is 0.20. How far does the block slide on the horizontal surface before it comes to rest?
In college softball, the distance from the pitcher's mound to the batter is 43 feet. If the ball leaves the bat at 110 mph , how much time elapses between the hit and the ball reaching the pitcher?
An electron enters a region of uniform electric field with an initial velocity of 50 km/s in the same direction as the electric field, which has magnitude E = 50 N/C, (a) what is the speed of the electron 1.5 ns after entering this region? (b) How far does the electron travel during the 1.5 ns interval?
A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

An athlete swings a 6.50-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.900 m at an angular speed of 0.700 rev/s. (a) What is the tangential speed of the ball

Answers

Answer:

v = 3.951 m/s

Explanation:

Given that,

Mass of a ball, m = 6.5 kg

Radius of the circle, r = 0.9 m

Angular speed of the ball, $\omega=0.7\ rev/s=4.39\ rad/s$

Let v is the tangential speed of the ball. It is given in terms of angular speed is follows :

$v=r\omega\n\nv=0.9* 4.39\n\nv=3.951\ m/s$

So, the tangential speed of the ball is 3.951 m/s.

6. A barber raises his customer's chair by applying a force of 150 N to thehydraulic piston of area 0.01 m2. If the chair is attached to a piston with an
area of 0.1 m², how much force is needed to raise the customer?
STEP 1: List the known
and unknown values F =
A=
A,
STEP 2: Write the
correct equation
STEP 3: Insert the
known values into the
equation to solve for
the unknown value

Answers

Answer:

15N

Explanation:

F¹=150N

A=0.01m2²

F2=?

A2=0.1m²

P=F/A

F1/A2=F2/A1

150/0.1=F2/0.01

Ugoing o pri
7.) True or False: "Courtney is traveled 5 miles in 3 hours" is an example of
acceleration.
True
False

Answers

True tell me if im correct

A sound wave travels with a velocity of 330 m/s and has a frequency of 500 Hz. What is itswavelength?

Answers

Answer:

Wavelength = 0.66 meters

Explanation:

Given the following data;

Speed = 330 m/s

Frequency = 500 Hz

To find the wavelength;

Mathematically, wavelength is calculated using this formula;

$Wavelength = \frac {speed}{frequency}$

Substituting into the equation, we have;

$Wavelength = \frac {330}{500}$

Wavelength = 0.66 meters

A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between ladder and floor is 0.40. What minimum angle can the ladder make with the floor before it slips?

Answers

The minimum angle that the ladder make with the floor before it slips is 51.34 Degree.

Given data:

The weight of ladder is, W = 100 N.

The length of ladder is, L = 8.0 m.

The coefficient of static friction between ladder and floor is, $\mu =0.40$ .

Apply the Newton' law in vertical direction to obtain the value of Normal Force (P) as,

$N = mg$

And force along the horizontal direction is,

$F= \mu * N\n\nF = \mu * mg$

Now, taking the torque along the either end of ladder as,

$-mgcos \theta * (L)/(2)+Fsin \theta * L =0\n\nmgcos \theta * (L)/(2) = Fsin \theta * L$

Solving as,

$mgcos \theta * (L)/(2) = (\mu mg) * sin \theta * L\n\ncos \theta * (1)/(2) = (\mu) * sin \theta\n\ntan \theta = (1)/(2 * 0.40 )\n\n\theta = tan^(-1)(1/0.80)\n\n\theta = 51.34^(\circ)$

Thus, we can conclude that the minimum angle that the ladder make with the floor before it slips is 51.34 Degree.

Learn more about the frictional force here:

brainly.com/question/4230804

Answer:

The minimum angle is 51.34°

Explanation:

Given that,

Weight of ladder = 100 N

Length = 8.0 m

Coefficient of static friction = 0.40

We need to calculate the normal force

Using Newtons law in vertical direction

$F_(y)=n-mg$

$N-mg=0$

$N=mg$

We need to calculate the normal force

Using Newtons law in horizontal direction

$F_(s)=f_(s)-P$

$f_(s)-P=0$

$f_(s)=P$

$P=\mu mg$

We need to calculate the minimum angle

Using torque about the point A then

$-mg\cos\theta* AB+P\sin\theta* AC=0$

Put the value into the formula

$mg\cos\theta*((L)/(2))=\mu mg\sin\theta* L$

$\cos\theta*(1)/(2)=\mu\sin\theta$

$(1)/(2)=\mu*\tan\theta$

$\theta=\tan^(-1)((1)/(2*0.40))$

$\theta=51.34^(\circ)$

Hence, The minimum angle is 51.34°

across a rough, horizontal surface. The chair's mass is 18.8 kg. The force you exert on the chair is 165 N directed 26 degrees below the horizontal. While you slide the chair a distance of 6.00 m , the chair's speed changes from 1.30 m/s to 2.50 m/s . Find the work done by friction on the chair.

Answers

Answer:

-847.2J

Explanation:

First find the acceleration from v^2= u^2 + 2as

v= 2.5 m/s

u= 1.3 m/s

a???

s=6.00

a= v^2-u^2/2s

a= (2.5)^2-(1.3)^2/2× 6

a= 0.38ms^-2

From Newtons second law:

(Force applied cos Θ) - (Frictional force) = ma

Frictional force = ma- (Force applied cos Θ)

Frictional force= (18.8×0.38) - (165 cos 26°)

Frictional force= 7.144- 148.3= -141.2N

Therefore,

Work done by friction = Frictional force × distance covered

= -141.2N × 6= -847.2J

Answer:

W = –847J

Explanation:

Given m = 18.8kg, F = 165N, θ = -26° (below the horizontal, s = 6.0m, u = 1.30m/s and v = 2.50m/s

In this problem, two forces act on the chair; the forward force F and the frictional force f. We would apply newton's second law to find the frictional force f after which we can calculate the workdone by the frictional force f×s.

But for us to apply newton's second law, we need to know the acceleration of the chair cause by the net force.

From constant acceleration motion equations

v² = u² + 2as

2.5² = 1.30² + 2a×6

6.25 = 1.69 +12a

12a = 6.25 – 1.69

12a = 4.56a

a = 4.56/12

a = 0.38m/s

By newton's second law the net sum of forces equals m×a

The force F has horizontal and vertical and components. It is the horizontal component of this force that pushes the chair against friction.

Fx and f are oppositely directed.

Fx – f =ma

165cos(-26) – f = 18.8×0.38

148.3 – f = 7.14

f = 148.3 – 7.14

f = 141.2N

Workdone = -fs = –141.2×6.00 = –847J

W = –847J

Work is negative because it is done by a force acting on the chair in a direction opposite (antiparallel) to that of the intended motion.

Other Questions

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.55 m and the acceleration achieved during the time the jumper is extending their legs is 1.2 times the acceleration due to gravity, g .How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

HELP ASAP PLEASE!!!In which direction(s) does the ground shake during an earthquake?A. sidewaysB. up and downC. back and forthD. all of the above

A radioactive nucleus has a half-life of 5*108 years. Assuming that a sample of rock (say, in an asteroid) solidified right after the solar system formed, approximately what fraction of the radioactive element should be left in the rock today? The age of the solar system is 4.5*109 years. (No calculator is necessary, but if it helps use it.)

When a box is placed on an inclined surface with no friction, it will: