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katrin [286]
2 years ago
13

Which of the following substances could be heated from 20°C to 100°C most quickly? Assume identical heat sources are used on all

substances.
Physics
2 answers:
Alex73 [517]2 years ago
7 0
It gold I think ...........
Lera25 [3.4K]2 years ago
3 0
Hello There!

Your answer is Gold because it has the lowest SH

Hope This Helps You!
Good Luck :) 

- Hannah ❤
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A cube with 30-cm-long sides is sitting on the bottom of an aquarium in which the water is one meter deep. (Round your answers t
OleMash [197]

Answer:

A) hydrostatic force on top of cube = 882.9N

B) hydrostatic force on sides of cube = 0N

Explanation:

Detailed explanation and calculation is shown in the image below

4 0
3 years ago
A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A
zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

V_{orbital} = \sqrt{\frac{GM_E}{R}}

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}

V_{orbital} = 5591.62m/s

V_{orbital} = 5.591*10^3m/s

PART B) The period of satellite is given as,

T = 2\pi \sqrt{\frac{r^3}{Gm_E}}

T = \frac{2\pi r}{V_{orbital}}

T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}

T = 238.61min

PART C) The gravitational force on the satellite is given by,

F = ma

F = \frac{1}{4} mg

F = \frac{270*9.8}{4}

F = 661.5N

5 0
3 years ago
1+5-(-5 - 1) whats the answer​
vladimir2022 [97]

Answer:

12

Explanation:

6 0
3 years ago
Read 2 more answers
Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic
Gre4nikov [31]

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

Let the weight that is hooked to two springs be w.

Spring#1:

Force constant= k

let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by \frac{1}{2}kx^{2} where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is, \frac{1}{2}k(x1)^{2}

                                                              ⇒\frac{1}{2}k(\frac{w}{k})^{2}

                                                              ⇒\frac{w^{2}}{2k}

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is, \frac{1}{2}2k(x2)^{2}

                                                              ⇒\frac{1}{2}2k(\frac{w}{2k})^{2}

                                                              ⇒\frac{w^{2}}{4k}

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is \frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=2:1

4 0
3 years ago
A 0.144-kg baseball is moving toward home plate with a speed of 43 m/s when
algol [13]
I would say 648858. bc yes
4 0
3 years ago
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