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tensa zangetsu [6.8K]
2 years ago
11

Draw suitable diagrams where necessary

Physics
1 answer:
mariarad [96]2 years ago
8 0

The experiment is described below in the explanation.

<h3>What is specific heat?</h3>

The specific heat is the amount of heat required to change the temperature by 1°C. It is denoted by C.

Heat lost or gained is represented as

Q = m C ΔT

In a calorimeter, a liquid is filled into the the cup with a temperature. A rod with high temperature is inserted into that liquid.

We use principle of  mixture by putting a heated rod in a liquid and then measuring the rise of temperature of liquid.

So, heat is lost by rod and gained by liquid and colorimeter.

Thus, experiment is described.

Learn more about specific heat.

brainly.com/question/11297584

#SPJ1

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A single, nonconstant force acts in the + x ‑direction on an object of mass M that is constrained to move along the x ‑axis. As
klasskru [66]

Answer:

3MQRT^2+\frac{9}{2}MR^2T^4

Explanation:

In order to find the work done by the force, we use the work-energy theorem, which states that the work done by a force on an object is equal to the change in kinetic energy of the object. Mathematically:

W=K_f-K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

W is the work done

m is the mass of the object

v is the final speed of the object

u is the initial speed

In this problem, we have:

m = M is the mass of the object

The position of the object at time t is

x(t)=P+Qt+Rt^3

We can find its speed at time t by calculating the derivative of the position:

v(t)=x'(t)=Q+3Rt^2

Therefore:

- The speed at time t = 0 is

u=v(0)=Q

- The speed at time t = T is

v=v(T)=Q+3RT^2

Substituting into eq.(1), we find the work done:

W=\frac{1}{2}M(Q+3RT^2)^2-\frac{1}{2}MQ^2=\\=\frac{1}{2}MQ^2+3QRT^2+\frac{9}{2}MR^2T^4-\frac{1}{2}MQ^2=\\=3MQRT^2+\frac{9}{2}MR^2T^4

6 0
3 years ago
A particle with charge +6.70 nC is in a uniform electric field directed to the left. Another force, in addition to the electric
mina [271]

Answer:

a) 4.6*10^{-5} J

b) 6133.33 J/C

c) 94358.9 C

Explanation:

(a) We have that the change in the kinetic energy equals the net work over the charge. Hence we have

\Delta E_K=W-W_E

where Ek is the kinetic energy, W is the work of the external force and WE is the work done by the electric field. By replacing we obtain:

W_E=W-\Delta E_k=7.60*10^{-5}J-3*10^{-5}J=4.6*10^{-5}J

(b) The potential difference is computed by using:

\Delta V=\frac{W_E}{q}=\frac{4.6*10^{-5}J}{7.5*10^{-9}C}=6133.33\frac{J}{C}

(c) With the work done by the electric force we can calculate the Electric field. By using the following formula we obtain:

W_E=qEd\\\\E=\frac{W_E}{qd}=\frac{4.6*10^{-5}J}{(7.5*10^{-9}C)(6.50*10^{-2}m)}=94358.9\frac{N}{C}

where we have used d=6.5cm=6.5*10^-9m and q=7.5nC=7.5*10^-9C

hope this helps!!

4 0
3 years ago
Read 2 more answers
HElP thank you pLeAsE
olga nikolaevna [1]

Answer:

Cell membrane

Explanation:

I've done this before

3 0
3 years ago
Read 2 more answers
A sonar wave is reflected from the ocean floor. For which angles of incidence do the wave’s angle of reflection equal its angle
nadya68 [22]
After a thorough research, there exists the same question that has the following choices.

<span>(1) angles less than 45°, only
(2) an angle of 45°, only
(3) angles greater than 45°, only
<span>(4) all angles of incidence

The correct answer (4) all angles of incidence.</span></span>
7 0
3 years ago
The force that moving, charged particles exert on one another is called
soldi70 [24.7K]
Electromagnetic force.

Hope I helped :-)
6 0
3 years ago
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