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cluponka [151]
3 years ago
11

2. If you are 5'10" tall, that is, 5 feet 10 inches, what is your height in meters? (2 54 cm = 1.00 in)

Physics
2 answers:
likoan [24]3 years ago
8 0

Answer:D 1.7

Explanation:

Just trust me

liraira [26]3 years ago
5 0

Answer:

5’10” = 1.778 = 1.8m

D

Explanation:

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Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter
Aleonysh [2.5K]

We can solve the problem by using the first law of thermodynamics:

\Delta U= Q-W

where

\Delta U is the variation of internal energy of the system

Q is the heat added to the system

W is the work done by the system

In this problem, the variation of internal energy of the system is

\Delta U=U_f-U_i=134 J-87 J=47 J

While the heat added to the system is

Q=500 J

therefore, the work done by the system is

W=Q-\Delta U=500 J-47 J=453 J

5 0
3 years ago
Read 2 more answers
A Carnot engine has an efficiency of 0.537, and the temperature of its cold reservoir is 379 K.
katen-ka-za [31]

Answer:

(A) Th = 818.6 K

(B) Qh = 14211.7 J

Explanation:

efficiency (n) = 0.537

temperature of cold reservoir (Tc) = 379 K

heat rejected (Qc) = 6580 J

(A) find the temperature of the hot reservoir (Th)

 n = 1 - \frac{Tc}{Th}

0.537 = 1 - \frac{379}{Th}

\frac{379}{Th} = 1 - 0.537 = 0.463        

Th = \frac{379}{0.463}

Th = 818.6 K

(B) what amount of heat is put into the engine (Qh) ?

from \frac{Tc}{Th} = \frac{Qc}{Qh}

Qh = 6580 ÷ \frac{379}{818.6}

Qh = 14211.7 J

8 0
3 years ago
John is traveling north at 20 meters/second and his friend Betty is traveling south at 20 meters/second. If north is the positiv
DanielleElmas [232]
John's velocity is 20 m/s north.
Betty's velocity is 20 m/s south.
Their speeds are both 20 m/s.
3 0
3 years ago
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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

4 0
3 years ago
What is concave mirror​
Licemer1 [7]

Answer:

A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex or concave. Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.

Explanation:

8 0
4 years ago
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