1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Mekhanik [1.2K]
2 years ago
12

Please I need help! This is the last question I need for this assignment!

Physics
1 answer:
Ne4ueva [31]2 years ago
6 0

Answer:

When the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.

What is specific heat capacity?

Specific heat capacity is the quantity

of heat required to raise a unit mass of

a substance by 1 kelvin.

Specific heat capacity of water and sand

{<em>refer to the above attachment}</em>

Δθ = Q/mc

Thus, for an equal mass of water and sand, when the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.

You might be interested in
Which of the following is a transformer used for?
mamaluj [8]

A transformer is used to increase or decrease a voltage.  BUT ... it has to be an AC voltage, or the transformer doesn't work.

5 0
3 years ago
Read 2 more answers
The ozone layer protects us from the harmful effects of which type of radiation?
leonid [27]

Answer:

Hey there

The ozone layer or ozone shield is a region of Earth's stratosphere that absorbs most of the Sun's ultraviolet radiation

Can u have brainly

3 0
2 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
A 120.0 kg crate is placed on a 15.00°
Citrus2011 [14]

F = 2820.1 N

Explanation:

Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as

Fnet = ma = 0 (a = 0 no sliding)

= F - mgsin15°

= 0

or

F = mgsin15°

= (120 kg)(9.8 m/s^2)sin15°

= 2820.1 N

7 0
3 years ago
The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a
s344n2d4d5 [400]

Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = 0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg

r = Radius of orbit of planet = 0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m

v = Orbital speed

The kinetic and potential energy balance is given by

\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days  

The orbital period is 14.45 days

5 0
3 years ago
Other questions:
  • A can of sardines is made to move along an x axis from x = 0.47 m to x = 1.20 m by a force with a magnitude given by F = exp(–8x
    11·1 answer
  • Can someone help me ASAP
    8·1 answer
  • Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o
    13·1 answer
  • How are speed and velocity similar ?
    9·2 answers
  • What are 3 facts about cnidarians?
    8·2 answers
  • An aluminum alloy rod has a length of 6.3243 cm at 16.00°C and a length of 6.3568 cm at the boiling point of water. (a) What is
    15·1 answer
  • the goat weighs 900 N and is 1 meter from the fulcrum. The strongman pulls down on the lever 3 meters from the fulcrum. What is
    8·1 answer
  • Three identical point charges of 2.0 μC are placed on the x-axis. The first charge is at the origin, the second to the right at
    5·1 answer
  • Hai vận động viên chạy trên cùng 1 đoạn đường, vận động
    10·1 answer
  • Is Cardiovascular Fitness the same as Anaerobic Exercise/Fitness?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!