How long does it take a cup of fresh water to freeze, boil and evaporate compared to fresh water

A projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall. (a) Determine the time necessary for the

jeka94

(a) 3.35 s

The time needed for the projectile to reach the ground depends only on the vertical motion of the projectile, which is a uniformly accelerated motion with constant acceleration

a = g = -9.8 m/s^2

towards the ground.

The initial height of the projectile is

h = 55.0 m

The vertical position of the projectile at time t is

$y = h + \frac{1}{2}at^2$

By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:

$0 = h + \frac{1}{2}at^2\\t=\sqrt{-\frac{2h}{a}}=\sqrt{-\frac{2(55.0 m)}{(-9.8 m/s^2)}}=3.35 s$

(b) 78.4 m

The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.

The horizontal motion is a uniform motion with constant velocity -

The horizontal velocity of the projectile is

$v_x = 23.4 m/s$

the time it takes the projectile to reach the ground is

t = 3.35 s

So, the horizontal distance covered by the projectile is

$d=v_x t = (23.4 m/s)(3.35 s)=78.4 m$

(c) 23.4 m/s, -32.8 m/s

The motion of the projectile consists of two independent motions:

- Along the horizontal direction, it is a uniform motion, so the horizontal velocity is always constant and it is equal to

$v_x = 23.4 m/s$

so this value is also the value of the horizontal velocity just before the projectile reaches the ground.

- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by

$v_y = u_y +at$

where

$u_y = 0$ is the initial vertical velocity

Using

a = g = -9.8 m/s^2

and

t = 3.35 s

We find the vertical velocity of the projectile just before reaching the ground

$v_y = 0 + (-9.8 m/s^2)(3.35 s)=-32.8 m/s$

and the negative sign means it points downward.

3 0

3 years ago

What was the position of the car at clock tick #3? At clock tick #5?

valkas [14]

Dbdbdnejdxjejejdjchfjdjdjdjdejejs

5 0

3 years ago

What would happen if you use a thicker wire around the iron nail of an electromagnet? (thats the whole question)

puteri [66]

Answer:

When we have a current I, we will have a magnetic field perpendicular to this current.

Then if we have a wire in a "spring" form. then we will have a magnetic field along the center of this "spring".

Now suppose we put an iron object in the middle (where the magnetic field is) then we will magnetize the iron object.

Of course, the intensity of the magnetic field is proportional to the current, given by:

B = (μ*I)/(2*π*r)

Where:

μ is a constant, I is the current and r is the distance between to the current.

Now remember that for a resistor:

R = ρ*L/A

R is the resistance, ρ is the resistivity, which depends on the material of the wire, L is the length of the wire, and A is the cross-section of the wire.

If we increase the area of the wire (if we use a thicker wire).

And the relation between resistance and current is:

I = V/R

Where V is the voltaje.

Now, if we use a thicker wire, then the cross-section area of the wire increases.

Notice in the resistance equation, that the cross-section area is on the denominator, then if we increase the area A, the resistance decreases.

And the resistance is on the denominator of the current equation, then if we decrease R, the current increases.

If the current increases, the magnetic field increases, which means that we will have a stronger electromagnet.

3 0

3 years ago

A mass movement that involves the sudden movement of a block of material is called a

Margarita [4]

A mass movement that involves the sudden movement of a block of material is called a B.slide.

<span>♡♡Hope I helped!!! :)♡♡
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7 0

3 years ago

Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la

rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93 ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

$v=v_0-g \,t$

for the object's vertical displacement we have

$y-y_0=v_0\,t - \frac{g}{2} \,t^2$

If the maximum height reached by the object is given in meters, we use the value for g in $m/s^2$ which is: $9.8\,\,m/s^2$

If the maximum height of the object is given in feet, we use the value for g in $ft/s^2$ which is : $32\,\,ft/s^2$

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

$v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}$

and now we use this to express the maximum height in the second equation we typed:

$y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}$

Then if the max height is 1.1 meters, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s$

If the max height is 1.1 feet, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s$

5 0

3 years ago

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