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2 years ago

Pamela has three computers, all of

Physics

2 answers:

Lena [83]2 years ago

6 0

Answer:

Pamela has three computers, all of them are very old.

Explanation:

Fiesta28 [93]2 years ago

5 0

U need to write better u egg head kid or tell your mom

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A golfer hits a golf ball at an angle of 25.0Â° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is

kvasek [131]

Answer:

Maximum height reached = 35.15 meter.

Explanation:

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

$0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}$

In the give problem we have R = 301.5 m, θ = 25° we need to find H.

So $\frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g$

Now we have $H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m$

So maximum height reached = 35.15 meter.

7 0

2 years ago

. 30

schepotkina [342]

Answer:

Explanation:

Length if the bar is 1m=100cm

The tip of the bar serves as fulcrum

A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum

The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.

Using moment

Sum of the moments of all forces about any point in the plane must be zero.

Let take moment about the fulcrum

100×20-F×2=0

2000-2F=0

2F=2000

Then, F=1000N

The force acting in the crate lid is 1000N

Option D is correct

7 0

2 years ago

In a harbor, you can see sea waves traveling around the edges of small stationary boats. Why does this happen?

faust18 [17]

Below are the choices that can be found in the other sources:

A. diffraction
B. refraction 
C. reflection 
D. transmission

The answer is diffraction. It means that the process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the wave forms produced.

8 0

2 years ago

The drawing shows the top view of a door that is 1.68 m wide. two forces are applied to the door as indicated. what is the magni

jekas [21]

First, torque is equal to force times the distance. for the first force that is applied, the torque is zero because is applied at the hinge. so the net torque:
t = ( 12 N ) ( 0 m ) ( cos 30 ) + ( 12 N ) ( 1.68 m ) cos 45
t = 14.26 Nm is the torque with respect to the hinge

8 0

2 years ago

A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the

tankabanditka [31]

Answer

given,

$F_L= 1.52\ N$