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3 years ago

Using the diagram below, calculate the PE and KE of the ball at the top, middle, and bottom of a drop. SHOW YOUR WORK!!

Physics

1 answer:

Orlov [11]3 years ago

5 0

Answer:

At the top

K.E = 0

P.E = 981 J

Mechanical energy is conserved

P.E is maximum

K.E = 0

At the middle

K.E = 981 J

P.E = 0

Mechanical energy is conserved

K.E is maximum

P.E = 0

At the bottom

K.E = 0

P.E = 981 J

Mechanical energy is conserved

K.E = 0

P.E is maximum

Explanation:

Given that

M= 5kg

H = 20 m

t = 7 s

According to conservation of energy,

At the top

K.E = 0

P.E = mgh

P.E = 5 × 9.81 × 20

P.E = 981 J

Mechanical energy is conserved

P.E is maximum

K.E = 0

At the middle

K.E = 981 J

P.E = 0

Mechanical energy is conserved

K.E is maximum

P.E = 0

At the bottom

K.E = 0

P.E = 981 J

Mechanical energy is conserved

K.E = 0

P.E is maximum

Given that

M= 5kg

H = 20 m

t = 7 s

At the top

K.E = 0

P.E = mgh

P.E = 5 × 9.81 × 20

P.E = 981 J

Mechanical energy is conserved

P.E is maximum

K.E = 0

At the middle

K.E = 981 J

P.E = 0

Mechanical energy is conserved

K.E is maximum

P.E = 0

At the bottom

K.E = 0

P.E = 981 J

Mechanical energy is conserved

K.E = 0

P.E is maximum

Explanation:

Given that

M= 5kg

H = 20 m

t = 7 s

At the top

K.E = 0

P.E = mgh

P.E = 5 × 9.81 × 20

P.E = 981 J

Mechanical energy is conserved

P.E is maximum

K.E = 0

At the middle

K.E = 981 J

P.E = 0

Mechanical energy is conserved

K.E is maximum

P.E = 0

At the bottom

K.E = 0

P.E = 981 J

Mechanical energy is conserved

K.E = 0

P.E is maximum

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A 36.3 kg cart has a velocity of 3 m/s. How much kinetic energy does the object have?

uysha [10]

Answer:

163.35

__________________________________________________________

<u>We are given:</u>

Mass of the object (m) = 36.3 kg

Velocity of the object (v) = 3 m/s

<u>Kinetic Energy of the object:</u>

We know that:

Kinetic Energy = 1/2(mv²)

KE = 1/2(36.3)(3)² [replacing the variables with the given values]

KE = 18.15 * 9

KE = 163.35 Joules

Hence, the cart has a Kinetic Energy of 163.35 Joules

7 0

3 years ago

The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensit

sveticcg [70]

Answer:

a) I = 3.63 W / m² , b) I = 0.750 W / m²

Explanation:

The intensity of a sound wave is given by the relation

I = P / A = ½ ρ v (2π f $s_{max}$ )²

I = (½ ρ v 4π² s_{max}²) f²

a) with the initial condition let's call the intensity Io

cte = (½ ρ v 4π² s_{max}²)

I₀ = cte s² f₀²

I₀ = cte 10 6

If frequency is increase f = 2.20 10³ Hz

I = constant (2.20 10³) 2

I = cte 4.84 10⁶

let's find the relationship of the two quantities

I / Io = 4.84

I = 4.84 Io

I = 4.84 0.750

I = 3.63 W / m²

b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or

I = cte (f s)²

I = constant (0.250 10³ 4)²

I = cte 1 10⁶

the relationship

I / Io = 1

I = Io

I = 0.750 W / m²

6 0

3 years ago

PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP

aleksley [76]

Answer:

red is not at the bottom. its density is 0.9 not 9!

also I don't need the brainlest thanks

6 0

3 years ago

Read 2 more answers

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

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3 years ago

What do you think causes the rising and sinking of molten rock in the mantle?

Svet_ta [14]

Answer:

Heyyy hope this helps

Convection currents describe the rising, spread, and sinking of gas, liquid, or molten material caused by the application of heat.

3 0

2 years ago