Gravity pushes objects away from each other TRUE OR FALSE

A rough surface for infrared waves may be polished for

zloy xaker [14]

Answer:

Radio Waves

Explanation:

7 0

3 years ago

A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 3300 kcal o

Tema [17]

Answer

given,

heat added to the gas,Q = 3300 kcal

initial volume, V₁ = 13.7 m³

final volume, V₂ = 19.7 m³

atmospheric pressure, P = 1.013 x 10⁵ Pa

a) Work done by the gas

W = P Δ V

W = 1.013 x 10⁵ x (19.7 - 13.7)

W = 6.029 x 10⁵ J

b) internal energy of the gas = ?

now,

change in internal energy

Δ U = Q - W

Q = 3300 x 10³ cal

Q = 3300 x 10³ x 4.186 J

Q = 1.38 x 10⁷ J

now,

Δ U = 1.38 x 10⁷ - 6.029 x 10⁵

Δ U = 1.32 x 10⁷ J

6 0

3 years ago

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

4 years ago

Pls someone I need it urgently and explain Solving and explanation so I can understand Thank you

Temka [501]

Answer:

f = 6.37 Hz, T = 0.157 s

Explanation:

The expression you have is

y = 5 sin (3x - 40t)

this is the equation of a traveling wave, the general form of the expression is

y = A sin (kx - wt)

where A is the amplitude of the motion, k the wave vector and w the angular velocity

Angle velocity and frequency are related

w = 2π f

f = w / 2π

from the equation w = 40 rad / s

f = 40 / 2π

f = 6.37 Hz

frequency and period are related

f = 1 / T

T = 1 / f

T = 1 / 6.37

T = 0.157 s

4 0

3 years ago

3 bulbs are in series and the same 3 bulbs are in parallel with the same battery. Which bulbs will be dimmer?

Marina86 [1]

3 bulbs are in series and if the same 3 bulbs are in parallel with the same battery then the bulbs that are connected in parallel will be dimmer

<h3>What is power?</h3>

The rate of doing work is known as power. The Si unit of power is the watt.

Power =work/time

The mathematical expression for the electric power is as follows

P = VI

The same current flows through both bulbs when they are connected in series. A greater voltage drop across the bulb with the higher resistance will result in higher power dissipation and brightness. In the case of the parallel combination, the bulb will be dimmer

Thus, If the same three bulbs are connected in series and parallel with the same battery, the parallelly connected bulbs will be dimmer, therefore the correct option is A

Learn more about power from here

brainly.com/question/3854047

#SPJ1

3 0

2 years ago