What is the length of the x-component of the vector shown below?

Two balls, ball A and ball B, are dropped from the same height onto the same surface. If ball A rebounds to a higher height than

mamaluj [8]

Answer:

its b

Explanation:

3 0

2 years ago

PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is

IgorLugansk [536]

Answer:

The image height is 3.0 cm

Explanation:

Given;

object distance, $d_o$ = 15.0 cm

image distance, $d_i$ = 5.0 cm

height of the object, $h_o$ = 9.0 cm

height of the image, $h_i$ = ?

Apply lens equation;

$\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm$

Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.

4 0

3 years ago

The kinetic friction force between a 60.0-kg object and a horizontal surface is 50.0 N. If the initial speed of the object is 25

Vikki [24]

Answer:

375 m.

Explanation:

From the question,

Work done by the frictional force = Kinetic energy of the object

F×d = 1/2m(v²-u²)..................... Equation 1

Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.

Make d the subject of the equation.

d = 1/2m(v²-u²)/F.................. Equation 2

Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.

Note: If is negative because it tends to oppose the motion of the object.

Substitute into equation 2

d = 1/2(60)(0²-25²)/-50

d = 30(-625)/-50

d = -18750/-50

d = 375 m.

Hence the it will slide before coming to rest = 375 m

6 0

3 years ago

A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas

Nookie1986 [14]

The maximum amount of work performed is

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Explanation:

The efficiency of a real heat engine is given by the equation:

$\eta = 1-\frac{T_C}{T_H}$ (1)

where

$T_C$ is the temperature of the cold reservoir

$T_H$ is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

$\eta = \frac{W_{max}}{Q_H}$

where

$W_{max}$ is the maximum work done

$Q_H$ is the heat absorbed from the hot reservoir

$Q_H$ can be written as

$Q_H=W_{max}+Q_C$

where

$Q_C$ is the heat released to the cold reservoir

So the previous equation can be also written as

$\eta=\frac{W_{max}}{W_{max}+Q_C}$ (2)

By combining eq.(1) and (2) we get

$1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}$

And re-arranging the equation and solving for $W_{max}$ , we find

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Learn more about work and heat:

brainly.com/question/4759369

brainly.com/question/3063912

brainly.com/question/3564634

#LearnwithBrainly

8 0

3 years ago

The internuclear distance between two closest Ar atoms in solid argon is about 3.8 A. The polarizability of argon is 1.66e-30 m3

oee [108]

Answer:

83.72 K

Explanation:

$\alpha$ = Polarizability of argon = $1.66\times 10^{-30}\ m^3$

I = First ionization = 1521 kJ/mol

r = Distance between atoms = 3.8 A

R = Gas constant = 8.314 J/mol K

T = Boiling point

Potential energy due to dispersion of gas is given by

$P=-\frac{3}{4}\frac{\alpha^2I}{r^6}\\\Rightarrow P=-\frac{3}{4}\frac{(1.66\times 10^{-30})^2\times 1521\times 10^3}{(3.8\times 10^{-10})^6}\\\Rightarrow P=-1044.01\ J/mol$

Kinetic energy is given by

$K=\frac{3}{2}RT$

The potential and kinetic energy will balance each other

$P=\frac{3}{2}RT\\\Rightarrow 1.04401\times 10^{-33}=\frac{3}{2}RT\\\Rightarrow T=\frac{1044.01\times 2}{3\times 8.314}\\\Rightarrow T=83.72\ K$

The boiling point of argon is 83.72 K

7 0

3 years ago