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Kamila [148]
3 years ago
6

The strongman lifts the pig by pulling down at position 1. How will the distance that he pulls down compare to the distance that

the pig moves up?
Question options:


The distance that the pig moves up will be more than the distance that the strongman pulls down.


The distance that the pig moves up will be the same as the distance that the strongman pulls down.


The distance that the pig moves up will be less than the distance that the strongman pulls down.


It is impossible to tell from this picture.

Physics
2 answers:
Elena L [17]3 years ago
7 0

Answer:

i believe its the 3rd option

Explanation:

if he pulls it down the pig will be level or higher than he pull down

Gennadij [26K]3 years ago
5 0

Answer:

C

Explanation:

The distance that the pig moves up will be less than the distance that the strongman pulls down.

Explanation: The strongman is further from the fulcrum than the pig is, so the distance that he pulls down will be greater than the distance that the pig moves up.

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An ultracentrifuge accelerates from rest to 99300 rpm in 1.95 min. Randomized Variables f = 99300 rpm t = 1.95 min l = 9.4 cm
Levart [38]

Answer: i'm pretty sure its part B

7 0
4 years ago
AP PHYSICS SYSTEM OF EQUATIONS
Anon25 [30]

Walking at a speed of 2.1 m/s, in the first 2 s John would have walked

(2.1 m/s) (2 s) = 4.2 m

Take this point in time to be the starting point. Then John's distance from the starting line at time <em>t</em> after the first 2 s is

<em>J(t)</em> = 4.2 m + (2.1 m/s) <em>t</em>

while Ryan's position is

<em>R(t)</em> = 100 m - (1.8 m/s) <em>t</em>

where Ryan's velocity is negative because he is moving in the opposite direction.

(b) Solve for the time when they meet. This happens when <em>J(t)</em> = <em>R(t)</em> :

4.2 m + (2.1 m/s) <em>t</em> = 100 m - (1.8 m/s) <em>t</em>

(2.1 m/s) <em>t</em> + (1.8 m/s) <em>t</em> = 100 m - 4.2 m

(3.9 m/s) <em>t</em> = 95.8 m

<em>t</em> = (95.8 m) / (3.9 m/s) ≈ 24.6 s

(a) Evaluate either <em>J(t)</em> or <em>R(t)</em> at the time from part (b).

<em>J</em> (24.6 s) = 4.2 m + (2.1 m/s) (24.6 s) ≈ 55.8 m

8 0
3 years ago
A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in
Gekata [30.6K]

A.

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

Substituting the values of the variables into the  equation, we have

Q = mcΔT

Q = ρVcΔT

Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C

Q = 482630400 J

So, the energy of the hot water is 482630400 J

B.

The elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

Substituting the values of the variables into the equation, we have

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Learn more about heat energy here:

brainly.com/question/11961649

5 0
3 years ago
Calculate the work done by a 4.2 N force pushing a 450. g sandwich across a table 0.8 m wide.
gladu [14]

Answer:

3.36 Joules

Explanation:

Work done= Force × distance

F=4.2 N

distance= 0.8 m

work done= 4.2 × 0.8

work done= 3.36 Joules

3 0
3 years ago
A wave with a frequency of 60.0 Hz travels through rubber with a wavelength of .90 m. What is the speed of this wave?
gizmo_the_mogwai [7]

Answer:

Speed of wave 54 ms⁻¹.

Explanation:

Given data:

Frequency of wave = 60 Hz

Wavelength of wave = 0.90 m

Speed = ?

Solution:

Formula

speed = wavelength × frequency

Now we will put the values in formula.

v = f × λ

Hz =  s⁻¹

v = 60 s⁻¹ × 0.90 m

v = 54 m s⁻¹

5 0
3 years ago
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