All categories

3 years ago

Question 1

Physics

2 answers:

Strike441 [17]3 years ago

7 0

The answers are a b c a a b a

Zielflug [23.3K]3 years ago

7 0

1 B
2 D
3 a
4b
5c
6A
there ya go

You might be interested in

What is the power of a 650N force that moves an object 75cm in 0.63s?

Ratling [72]

Answer:

Power is 773.8 watts

Explanation:

${ \rm{power = \frac{force \times distance}{time} }} \\ \\ { \rm{power = \frac{650 \times 0.75}{0.63} }} \\ \\ { \rm{power = 773.8 \: watts}}$

5 0

2 years ago

Swings are usually made out of materials that do not conduct heat easily, so that children will not burn themselves as they play

Irina18 [472]

the correct answer is D: all of these

8 0

3 years ago

HELP PLEASED THIS IS DUE TODAY AND ALSO NO LINKS PLEASED ALSO FILES PLEASED AND THANKS>

Katyanochek1 [597]

1. 2 way radio

2. radio waves

3. communication

4. convert the voltage from a transmitter into a radio signal

5. signal strength refers to the transmitter power output

6. sorry, not so sure.. Though it might be the waves.

4 0

3 years ago

A Carnot engine whose high-temperature reservoir is at 464 K has an efficiency of 25.0%. By how much should the temperature of t

nexus9112 [7]

Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

$\epsilon = 1 - \dfrac{T_C}{T_H}$

$0.25 = 1 - \dfrac{T_C}{464}$

$T_C= (1 - 0.25) \times 464$

$T_C= 0.75 \times 464$

T_C = 348 K

now,

if the efficiency is equal to 42$ = 0.42

$T_C= (1 - 0.42) \times 464$

$T_C= 0.58 \times 464$

$T_C= 269.12\ K$

8 0

3 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago