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777dan777 [17]
3 years ago
11

Calculate a. The heat that must be supplied to a 500.0 g copper kettle containing 400.0 g of water to raise its temperature from

22.0°C to the boiling point of water, 100.0°C. b. What percentage of the heat is used to raise the temperature of the water?
Physics
1 answer:
uysha [10]3 years ago
7 0

Explanation:

<h2>For Copper</h2>

dH copper = mCdT copper

<em>(Specific Heat of copper=0.385 J/g C )   </em>

dH = 500 g (0.385 J/g C) (78 C rise)

dH = 15,015 Joules

<h2> For Water</h2>

dH water =  m C dT water

<em>(Specific Heat of copper=</em>4.184 J/g-C<em>)   </em>

<em />

dH = 400 g (4.184 J/g-C) (78 C rise)

dH = 130,540 Joules

total heat = 15,015 + 130,540 = 145,555 Joules

<h2>Percentage for Water  </h2>

(130,540 Joules  / 145,555 Joules) x 100 = 89.7 %

If we consider that we have 3 Significant Figures,

then, the answers become ,

15.0 KJ must be added for Copper

130.5 KJ must be added for Water

and the total of 145 KJ must be added in the kettle with the water

89.7 %  of heat goes to the Water

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A flute plays a note with a frequency of 266 Hz. What is the speed of sound if the wavelength is 1.3 m?
hjlf

Wave speed  =  (frequency)  x  (wavelength)

                       =  (266 /sec)  x  (1.3 meters)

                       =     345.8 meters/sec
6 0
3 years ago
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A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0=4\pi \cdot 10^{-7} H/m is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m

4 0
3 years ago
Please help me, this is a physics test.
sweet-ann [11.9K]

Answer:

a = 2 [m/s²]

Explanation:

To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.

We can use the following equation.

v_{f}=v_{o}+a*t\\

where:

Vf = final velocity = 11 [m/s]

Vo = initial velocity = 5 [m/s]

a = acceleration [m/s²]

t = time = 3 [s]

11 = 5 + a*3\\6=3*a\\a= 2[m/s^{2} ]

4 0
3 years ago
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by stri
gogolik [260]

Answer:

(a) 1 : 2

(b) same

Explanation:

Let the mass of puck A is m and the mass of puck B is 2 m.

initial speed for both the pucks is same as u and the distance is same for both is s.

let the tension is T for same.

The kinetic energy is given by

K = 0.5 mv^2

(a) As the speed is same, so the kinetic energy depends on the mass.

So, kinetic energy of A : Kinetic energy of B = m : 2m  = 1 : 2

(b) A the distance s same so the final velocities are also same.

8 0
2 years ago
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