Uneven surfaces show ---------- reflection

Assume that a friend hands you a 10-newton box to hold for her. If you hold the box without moving it at a height of 10 meters a

malfutka [58]

Answer:

100 Joules

Explanation:

Applying,

W = mgh................... Equation 1

Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.

From the question,

Given: mg = 10 newtons, h = 10 meters.

Substitute these values into equation 1

W = 10×10

W = 100 Joules.

Hence the amount of workdone is 100 Joules

7 0

3 years ago

Air expands isentropically from 2.2 MPa and 77°C to 0.4 MPa. Calculate the ratio of the initial to the final speed of sound.

djyliett [7]

Answer:

The ratio of initial to final speed of sound is given as 1.28.

Explanation:

As per the thermodynamic relation of isentropic expansion

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

Here

$P_1$ is the pressure at point 1 which is given as 2.2 MPa
$T_1$ is the temperature at point 1 which is given as 77 °C or 273+77=350K

$P_2$ is the pressure at point 1 which is given as 0.4 MPa
$T_2$ is the temperature at point 2 which is to be calculated
k is the ratio of specific heats given as 1.4

Substituting values in the equation

$\frac{T_2}{350}=(\frac{0.4}{2.2})^{\frac{1.4-1}{1.4}}\\\frac{T_2}{350}=(0.18)^{0.2857}\\T_2=(0.18)^{0.2857} \times 350 \\T_2=0.61266 \times 350\\T_2=214.43 K$

As speed of sound c is given as

$c=\sqrt{kRT}$

for initial to final values it is given as

$\frac{c_i}{c_f}=\frac{\sqrt{k_1R_1T_1}}{\sqrt{k_2R_2T_2}}$

As values of k and R is constant so the ratio is given as

$\frac{c_i}{c_f}=\sqrt{\frac{T_1}{T_2}}$

Substituting values give

$\frac{c_i}{c_f}=\sqrt{\frac{350}{214.43}}\\\frac{c_i}{c_f}=\sqrt{1.63}}\\\frac{c_i}{c_f}=1.277 \approx 1.28$

So the ratio of initial to final speed of sound is 1.28.

5 0

3 years ago

In the Fresnel circular aperture setup, the distances from the aperture to the light source and the reception screen are 1.5 m a

ivanzaharov [21]

Explanation:

The width of the central maximum is given by

W = 2 λ L / a

where W is the width of the central maximum

λ is the wavelength of the light used.

L is the distance between the aperture and screen

a is the width of the slit or aperture

So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.

5 0

3 years ago

Is a baseball and cannon are dropped from the same height at the same time, which ball will hit the ground first?

lara31 [8.8K]

If you include the effects of falling through air, then you have to know the
shape, size, weight, and surface texture of the objects. You also have to
know the height from which they're dropped, and the temperature, pressure,
and humidity of the air. All these things make a difference in how they fall.

If you ignore the effects of falling through air, like build a giant metal tank
and pump all the air out of it, and ONLY talk about the effects of gravity, then
ALL OBJECTS accelerate at the same rate. If you drop two things from the
same height at the same time, then they both hit the ground at the same time,
traveling at the same speed, no matter what they are. They could be a piece of
tissue and a car !

There are several museums where they have a big glass pipe that you can
see through, and they pump the air out of the pipe and drop a feather and a
bowling ball from the top inside at the same time, and they both reach the
bottom together.

If gravity is the only force on an object, then all objects fall at the same rate.

8 0

3 years ago

Read 2 more answers

A circular loop of radius 13 cm carries a current of 16 A. A flat coil of radius 0.63 cm, having 48 turns and a current of 1.5 A

azamat

Answer:

a) Bt = 7.73 * 10^-5 T

b) T = 6.94 * 10^-7 N*m

Explanation:

Step 1: Data given

Circumar loop Radius = 13 cm

Current = 16 A

Flat coil radius = 0.63 cm

48 turns

Current = 1.5 A

a) What is the magnitude of (a) the magnetic field produced by the loop at its center

Let's assume a loop concentric with a coil, the plane of the coil is perpendicular to the plane of the loop. The magnetic field due to the loop at the center of the loop can be given by:

Bt = µ0It / 2Rt

In this case we'll get:

Bt = ((4π * 10^-7 T*m/A)(16A)) /(2*0.13m)

Bt = 7.73 * 10^-5 T

 b) What is the magnitude of the torque on the coil due to the loop?

The torque magnitude excreting on the coil due to the magnetic field of the loop is given by:

T = µcBtsin(∅)

with µc = the magnetic dipole moment of the coil

with ∅ = the angle between the magnetic dipole moment and the magnetic field. The magnetic dipole moment is given by:

µc = N*Ic*A

⇒ with N = the number of turns in the coil

⇒ with A = πRc² = the area of the coil

µc =π*N*Ic*Rc²

T= π*N*Ic*Rc²*Bt(sin∅)

In this situation we'll have:

T= π*48*1.5A* (0.63 *10^-2m)²*(7.73 * 10^-5 T)*sin(90)

T = 6.94 * 10^-7 N*m

8 0

3 years ago

Uneven surfaces show ----------reflection ​

Uneven surfaces show ----------reflection