All categories

3 years ago

Four weightlifters (A-D) enter a competition. The mass, distance, and time of their lifts are shown in the table.

Physics

1 answer:

siniylev [52]3 years ago

7 0

Let Pa, Pb, Pc, and Pd be the powers delivered by weightlifters A, B, C, and D, respectively.

Use this equation to determine each power value:

P = W÷Δt

P is the power, W is the work done by the weightlifter, and Δt is the elapsed time.

A) Determining Pa:

Pa = W÷Δt

The weightlifter does work to lift the weight up a certain distance. Therefore the work done is equal to the weight's gain in gravitational potential energy. The equation for gravitational PE is

PE = mgh

PE is the potential energy, m is the mass of the weight, g is the acceleration of objects due to earth's gravity, and h is the distance the weight was lifted.

We can equate W = PE = mgh, therefore we can make the following substitution:

Pa = mgh÷Δt

Given values:

m = 100.0kg

g = 9.81m/s²

h = 2.25m

Δt = 0.151s

Plug in the values and solve for Pa

Pa = 100.0×9.81×2.25÷0.151

Pa = 14600W (watt is the SI derived unit of power)

B) Determining Pb:

Let us use our new equation derived in part A to solve for Pb:

Pb = mgh÷Δt

Given values:

m = 150.0kg

g = 9.81m/s²

h = 1.76m

Δt = 0.052s

Plug in the values and solve for Pb

Pb = 150.0×9.81×1.76÷0.052

Pb = 49800W

C) Determining Pc:

Pc = mgh÷Δt

Given values:

m = 200.0kg

g = 9.81m/s²

h = 1.50m

Δt = 0.217s

Plug in the values and solve for Pc

Pc = 200.0×9.81×1.50÷0.217

Pc = 13600W

D) Determining Pd:

Pd = mgh÷Δt

Given values:

m = 250.0kg

g = 9.81m/s²

h = 1.25m

Δt = 0.206s

Plug in the values and solve for Pd

Pd = 250.0×9.81×1.25÷0.206

Pd = 14900W

Compare the following power values:

Pa = 14600W, Pb = 49800W, Pc = 13600W, Pd = 14900W

Pc is the lowest value.

Therefore, weightlifter C delivers the least power.

You might be interested in

List the steps you would use to reach one of your personal growth goals.

Zinaida [17]

Im not sure, but here are a few of mine : Learn to be silent, and listen, take massive action and be proactive, listen, focus, and lastly, persist. Hope this helped! (:

6 0

3 years ago

Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by

OlgaM077 [116]

Explanation:

Area of ring $\ 2{\pi} a d a$

Charge of on ring $d q=-(\ 2{\pi} a d a)$

Charge on disk

$Q=-\left(\pi R^{2}\right)$

$\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}$

Note: Refer the image attached

8 0

3 years ago

Two pulses are moving along a string one pulse is moving to the right and the second is moving to the left. Both pulses reach po

MatroZZZ [7]

Answer:

Constructive interference

Explanation:

- This is an example of a standing wave produced when two ends of a string are oscillated in the same plane. The displacement of of point on two ends oscillates vertically.

- We are given that two pulses move along the string each coming towards each other and meet at a common point ( P ).

- Each pulse have their own magnitude or displacement in the vertical plane. If the pulses are to meet at a common point at the same instant, then they interfere with each other constructively.

- Where constructive interference of two pulses is the addition of magnitudes of induvidual pulses and form a single puls of the constructed magnitude.

magnitude ( New pulse ) = magnitude (Pulse 1) + magnitude (Pulse 2)

6 0

3 years ago

Read 2 more answers

Four point charges, each of magnitude 2.38 µC, are placed at the corners of a square 75.2 cm on a side. If three of the charges

poizon [28]

Answer:

The Electric Force on Negative Charge is 2.968 N

Explanation:

charge on each corner, q = 2.38 micro coulomb

Side of square, a = 75.2 cm

Coulombic constant, K = 8.98755 x 10^9 Nm²/C²

sides of the square are A,B,C and D

and all sides of a square are equal so

AB = BC = CD = DA = 75.2 cm = 0.752 m

Diagonal, AC = BD = 1.414 x 0.752 = 1.06 m

Electric field at D due to charge at A

EA= Kq÷AB^2

= 8.98755×10^9 × 9.87×10^-6 ÷ 0.752^2

EA= 156863.82 N/C

Similarly Electric field at D due to charge C

EC=Kq÷CD^2

= 8.98755×10^9 ×9.87×10^-6 ÷ 0.752^2

EC= 156863.82 N/C

Electric field at D due to charge at BB

EB=Kq÷BD^2

EB=8.98755×10^9 × 9.87×10^-6 ÷ 1.06^2

EB=78949.01 N/C

Resolve the compoents

Ex = EA + EB cos 45

Ex = 156863.82 + 78949.01 x 0.707

Ex = 212689.2 N/C

Ey = EC + EB Sin 45

Ey = 156863.82 + 78949.01 x 0.707

Ey = 212689.2 N/C

The resultant electric field is

E = 1.414 x 212689.2 = 300787.95 N/C

the electric force on the negative charge is

F = q x E

F = 9.87 x 10^-6 x 300787.95

F = 2.968 N

7 0

2 years ago

A body builder loads a bar with 550 newtons (~125 pounds) of weight and pushes the bar over her head 10 times. Each time she lif

Neporo4naja [7]

Work = (force) x (distance)

(550 newtons) x (0.5 meter) = 275 joules each lift .

(275 joules/lift) x (10 lifts) = 2,750 joules of work all together.

Power = (work done) / (time to do the work)

= (2,750 joules) / (20 seconds)

= 137.5 watts . (about 0.18 horsepower)

6 0

3 years ago