How do you determine the specific heat of a substance?

DiKsa [7]

This has to do with independent and dependent variables. The independent variable is what affects the dependant variable, so the question is, does a students mark on a test affect their hours of sleep or does their hours of sleep affect their test results? Which one makes more sense? I would say the latter.

6 0

2 years ago

When a car driving up a hill with constant speed: I. its kinetic energy is decreasing. II. its potential energy is constant. III

madam [21]

A car driving up a hill at a constant speed experiences no change in its kinetic energy while it's potential energy increases with increasing height, thus none of the options are correct.

Understanding the concept

Consider a car moving up the hill at a constant speed as shown in the figure below. The following forces act on the car:

N is the normal reaction force acting in an upward direction

f_s is the static friction force exerted due to friction between the road and the tires of the car
f_k is the rolling friction force in the direction opposing that of the tire
mg is the force acting in a downward direction.

θ is the angle of inclination.

Here as the car is moving up the hill at a constant speed, the net force exerted on the car is zero. Also, the kinetic energy of the car will not change as its velocity is constant and the potential energy will change with increasing height. Thus, none of the given options are correct.

Learn more about motion on an incline here:

brainly.com/question/13513083

#SPJ4

5 0

1 year ago

An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e

timama [110]

Answer:

K = 80.75 MeV

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

$E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}$

where $E_{ph}$ : is the photon energy, $E_{0p} and E_{0ap}$ : are the rest energies of the proton and the antiproton, respectively, equals to m₀c², $K_{p} and K_{ap}$ : are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.

Therefore the kinetic energy of the antiproton is:

$K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}$

The proton mass is equal to the antiproton mass, so:

$K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV$

$K_{ap} = 80.75 MeV$

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

3 0

3 years ago

Use the drop-down menus to complete the statement.

VashaNatasha [74]

Answer: A is your answer i am sorry if i am wrong

Explanation:

he first PLCs were programmed with a technique that was based on relay logic wiring schematics. This eliminated the need to teach the electricians, technicians and engineers how to program a computer - but, this method has stuck and it is the most common technique for programming PLCs today.

7 0

3 years ago

Read 2 more answers

A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released

GREYUIT [131]

Answer:

$6.75\mu C/m^2$