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evablogger [386]
2 years ago
9

Four students measured the acceleration of gravity. The accepted value for their location is 9.78 m/s2. Which student's measurem

ent has the largest percent error?
Physics
2 answers:
garri49 [273]2 years ago
6 0
<span>the answer would actually be student 1 :9.87 m/s2, not 9.78</span>
Vitek1552 [10]2 years ago
4 0

Answer:

9.45m/s^2

Explanation:

I just took the quiz and if you have options "9.85, 9.45, 10.01, 9.75" 9.45m/s^2 is the correct option.

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A dipole of moment 0.5 e·nm is placed in a uniform electric field with a magnitude of 8 times 104 N/C. What is the magnitude of
Over [174]

Answer:

The net torque is zero

Explanation:

Let's assume that the dipole is compose of two equal but oposite charges e, and it cam be represented by a rod with one end having a charge e and the other end with a charge of -e. Notice that the dipole is parallel to the electric field thus the force felt by both of the charges will be parallel to the electric field. This means that there will be no components of the forces that are perpendicular to the rod which is a requirement for it to have a torque.

8 0
2 years ago
A ball is tossed in the air and released. It moves up, reverses direction, falls back down again, and is caught at the same heig
PolarNik [594]

Answer:

The potential energy has a  maximum when the ball is a time that is half of the time for total travel

Explanation:

Generally potential energy is a the varies directly with the height according to this formula

            PE =mgh

and the ball attains a maximum height when the time is equal to half of the total time taken to travel  

8 0
2 years ago
Read 2 more answers
The water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of
Damm [24]

Answer:

P_(pump) = 98,000 Pa

Explanation:

We are given;

h2 = 30m

h1 = 20m

Density; ρ = 1000 kg/m³

First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

Thus, it can be expressed as;

P_(tank)+ P_(pump) = P_(nozzle)

Now, the pressure would be given by;

P = ρgh

So,

ρgh_1 + P_(pump) = ρgh_2

Thus,

P_(pump) = ρg(h_2 - h_1)

Plugging in the relevant values to obtain;

P_(pump) = 1000•9.8(30 - 20)

P_(pump) = 98,000 Pa

5 0
2 years ago
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
1 year ago
A cement factory emits 900 kilograms of CO2 to produce 1,000 kilograms of cement. A fully grown tree removes six kilograms of CO
Dmitriy789 [7]
<h3><u>Answer;</u></h3>

To make the factory carbon neutral, the owners need to grow<em><u> 15000 trees</u></em> over an area of land measuring <em><u>75 acres</u></em>

<h3><u>Explanation and solution;</u></h3>

From the information;

900 kg CO2 = 1000 kg Cement

1 tree = 6 kg CO2

1 acre = 200 trees

<em>100000 kg Cement will require;</em>

<em>=(900 × 100000)/1000</em>

<em>= 90,000 kg of CO2</em>

<em>But 1 tree = 6 kg of CO2</em>

<em>Number of trees = 90,000/6</em>

<em>                            = 15,000 trees </em>

<em>But, 1 acre = 200 trees</em>

<em>Number of acres = 15,000/200</em>

<em>                             = 75 acres of land </em>

Therefore;

To make the factory carbon neutral, the owners need to grow<em><u> 15000 trees</u></em> over an area of land measuring <em><u>75 acres</u></em>

6 0
2 years ago
Read 2 more answers
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