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loris [4]
2 years ago
12

If 11,550 J heat is absorbed by a copper object when it is heated from 20.0°C to 30.0°C,

Physics
1 answer:
RSB [31]2 years ago
5 0

Answer:

The mass of the copper is 3.00kg

Explanation:

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ivanzaharov [21]

Answer:

(1) chess is played by him

(2) The great black engine was seen by her

(3) Arrangements for functions had been made by him

(4) A meeting is being attended by Raman

= when the last of the guests left, I went back into the hall to look for my mobile phone that I had kept on the table, I found that my mobile phone was not there. I felt freaked and I asked to every member of my family about my phone, but nobody had idea about that. then there was a phone call made by my uncle to my father, that his son had taken the mobile by mistake. After sometime, My uncle came to return my mobile. Then I took a sigh of relief .

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3 0
3 years ago
What is the resultant force acting on an object? The first force is 45 N West and the second is 23 N East.
Paul [167]

Answer:

22N West

Explanation:

45-23 because they are in opposite directions.

4 0
3 years ago
Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m
maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

T is the tension in the string

\mu is the mass per unit length

For the string in the problem,

L = 30.0 m

\mu=9.00\cdot 10^{-3} kg/m

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz

The next frequencies (harmonics) are given by

f_n = nf

with n being an integer number and f being the fundamental frequency.

So we get:

f_2 = 2 (0.786 Hz)=1.572 Hz

f_3 = 3 (0.786 Hz)=2.358 Hz

f_4 = 4 (0.786 Hz)=3.144 Hz

6 0
3 years ago
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6
lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

I = \frac{P_{avg}}{A}

I = \frac{P_{avg}}{4\pi r^2}

I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m

I = 6.529*10^{-6}W/m^2

The amplitude of electric field at the receiver is

I = \frac{E_{max}^2}{2\mu_0 c}

E_{max}= \sqrt{2I\mu_0 c}

The amplitude of induced emf by this signal between the ends of the receiving antenna is

\epsilon_{max} = E_{max} d

\epsilon_{max} = \sqrt{2I \mu_0 cd}

Here,

I = Current

\mu_0 = Permeability at free space

c = Light speed

d = Distance

Replacing,

\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}

\epsilon_{max} = 0.05434V

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0
3 years ago
A light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of
Dennis_Churaev [7]

Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = 85\frac{1000}{3600}=23.61\ m/s

Initial velocity = u = 0

Equation of motion

v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2

Emergency acceleration is 2.99 m/s² (magnitude)

6 0
2 years ago
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