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Lorico [155]
3 years ago
6

7.

Physics
1 answer:
jeka943 years ago
5 0

الملك - الإتاوات - المحاكم - المجلس الملكي - القانون الروماني

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True or False? PLEASE HELP ME​
viva [34]

Answer:

a. True - Joules is the unit measure for energy.

b. False - Potential energy is associated with position

c. False - Kinetic energy is associated with movement.

d. False - It's climbing, which means it also has kinetic energy.

6 0
3 years ago
Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro
Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

7 0
3 years ago
How large is the acceleration of a 25 kg mass with a net force of 75 N applied horizontally to it?
Sveta_85 [38]

Answer:

Explanation:

F = ma

a = F/m

a = 75/25

a = 3 m/s²

8 0
2 years ago
What would be the escape speed for a craft launched from a space elevator at a height of 54,000 km?
Natasha_Volkova [10]

Answer: 3.63 km/s

Explanation:

The escape velocity equation for a craft launched from the Earth surface is:

V_{e}=\sqrt{\frac{2GM}{R}}

Where:

V_{e} is the escape velocity

G=6.67(10)^{-11} Nm^{2}/kg^{2} is the Universal Gravitational constant

M=5.976(10)^{24}kg is the mass of the Earth

R=6371 km=6371000 m is the Earth's radius

However, in this situation the craft would be launched at a height h=54000 km=54000000 m over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

V_{e}=\sqrt{\frac{2GM}{R+h}}

V_{e}=\sqrt{\frac{2(6.67(10)^{-11} Nm^{2}/kg^{2})(5.976(10)^{24}kg)}{6371000 m+54000000 m}}

Finally:

V_{e}=3633.86 m/s \approx 3.63 km/s

7 0
3 years ago
A 108 kg clock initially at rest on a horizontal floor requires a 653 N horizontal force to set it in motion. After the clock is
kifflom [539]

Answer:

\mu_s=0.61

\mu_k=0.49

Explanation:

Given that,

Mass of the clock, m = 108 kg

When the clock is not moving, force acting on it, F_1=653\ N

For the clock in motion, force acting on it, F_2=527\ N

To find,

\mu_s\ and\ \mu_k

Solution,

When an object is at rest, the force acting on it is called force due to static friction and if the object is in motion, the force acting on its called force due to kinetic friction.

Let \mu_s is the coefficient of static friction. Force is given by :

F_s=\mu_smg

\mu_s=\dfrac{F_s}{mg}

\mu_s=\dfrac{653}{108\times 9.8}

\mu_s=0.61

Let  \mu_k is the coefficient of kinetic friction. Force is given by :

F_k=\mu_kmg

\mu_k=\dfrac{F_k}{mg}

\mu_k=\dfrac{527}{108\times 9.8}

\mu_k=0.49

8 0
3 years ago
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