All categories

3 years ago

What is brightness. help asap

Physics

1 answer:

notka56 [123]3 years ago

8 0

Answer:

The subjective visual sensation related to the intensity of light emanating from a surface or from a point source.

You might be interested in

A car is traveling at 39.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of fricti

777dan777 [17]

Answer:

The minimum distance in which the car will stop is

x=167.38m

Explanation:

$39.7\frac{mi}{h}*\frac{1km}{0.621371mi}*\frac{1000m}{1km}*\frac{1h}{3600s}=17.747\frac{m}{s}$

∑F=m*a

∑F=u*m*g

The force of friction is the same value but in different direction of the force moving the car so it can stop so

$F=m*a\\a=\frac{F}{m}\\a=\frac{u*m*g}{m}\\a=u*g\\a=0.096*-9.8\frac{m}{s^{2} }$

$a=-0.9408 \frac{m}{s^{2}}$

$v_{f}^{2}=v_{o}^{2}+2*a*(x_{f}-x_{o})\\v_{f}=0 \\x_{o}=0\\0=v_{o}^{2}+2*a*x_{f}\\x_{f}=\frac{v_{o}^{2}}{2*a} \\x_{f}=\frac{(-17.747\frac{m}{s})^{2}}{2*(-0.9408)} \\x_{f}=167.38m$

4 0

2 years ago

Why does increasing the number of trials increase confidence in the results of the experiment?

Paladinen [302]

It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.

8 0

3 years ago

Read 2 more answers

HELP!

Goryan [66]

Answer:

The solved problem is in the photo. Hope it helps.

3 0

3 years ago

An object that is slowing down in a positive direction must have

zepelin [54]

Answer:

Positive velocity and negative acceleration

Explanation:

An object moving in the positive direction has a positive velocity.

An object that's slowing down while moving in the positive direction has a negative acceleration.

5 0

3 years ago

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and

nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}$

k = 1.4

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K$

Work done is given as;

$W = \frac{1}{2} *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

6 0

3 years ago