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Anestetic [448]
3 years ago
12

How much heat in joules would you need to raise the temperature of 1 kg of water by 5 °C?

Physics
1 answer:
labwork [276]3 years ago
7 0
First we need to know the specific heat capacity of water. You can look it up online because it is uniform. The specific heat capacity of water is 4184 kg^-1 °C^-1.

To calculate the energy required to heat a specific thing up, we can use the formula
E = mc△T
Where E is the total energy required, m as mass, c as specific heat capacity and △T

Substitute the numbers in
E = 1 x 4184 x 5
E = 20920J

Therefore the answer is 20920 joules
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When a person wearing a helmet rides a bicycle, the helmet experiences which type of friction?
vodomira [7]

Answer:

B

Explanation:

3 0
3 years ago
Read 2 more answers
Drag the item from the item bank to its corresponding match.1.A fireman turns on his hose and is knocked backwards.2.It takes le
creativ13 [48]

Answer:

1 - third law

2 - second law

3 - first law

4 - third law

5 - second law

6 - first law

Explanation:

First law

In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.

Second law

In an inertial frame of reference, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration, a of the object

F = ma.

Third law

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

7 0
3 years ago
Sinusoidal waves 5.00 cm in amplitude are to be transmitted along a string that has a linear mass density of 4.00× 10⁻²kg /m . T
mylen [45]

The highest frequency (f) at which the source can operate is given as:

f = 55.133Hz.

<h3>What are sinusoidal waves?</h3>

The most realistic representation of how many objects in nature change state is a sine wave or sinusoidal wave.

A sine wave depicts how the intensity of a variable varies over time.

<h3>What is the calculation justifying the above result?</h3>

P = (1/2) μω²A²v

300W = 1/2 (4 X 10⁻²kg/m) ω₂ (0.05m)²v

Thus the wave speed is:

v = √(T/μ)

= √[(100N)/(4 X 10⁻²kg/m)

= 50m/s

300W = 1/2(4 X 10⁻²kg/m) ω²(0.05m)² (50m/s)

⇒ ω = 346.41 1/s

ω = 346.41 1/s

= 2πf

⇒ f = 55.133Hz


Learn more about Sinusoidal waves:
brainly.com/question/20912200
#SPJ4

7 0
2 years ago
A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hits the ground. From w
BaLLatris [955]

Answer:

The object fell from about 38.14 meters

Explanation:

We can use the formula for displacement under accelerated motion due to gravity to find the velocity of the object had 30 m before hitting the ground :

y_f-y_i=v_i * \,t-\frac{9.8}{2} \,t^2\\0-30=v_i \,(1.5)-4.9\,(1.5)^2\\-18.975=v_i\,(1.5)\\v_i=-12.65\,\,\frac{m}{s}

Now, knowing this velocity, we can find the time it took to fall from the initial position to 30 m before hitting the ground:

v_f=v_i-g\,*\,t\\-12.65=0-g \,*\,t\\t=\frac{12.65}{g} \\t\approx1.29\,\,s

And now we can find what is the total distance covered in 1.5 s plus 1.29 seconds for this free falling object:

y_f-y_i=v_i * \,t-\frac{9.8}{2} \,t^2\\0-y_i=0-\frac{9.8}{2} \,(2.79)^2\\-y_i=-38.14\\y_i=38.14\,\,m

7 0
2 years ago
The starship Enterprise leaves a distant earth colony, which is nearly at rest with respect to the earth. The ship travels at ne
slamgirl [31]

Explanation:

t = t/v(1-v^2/c^2)

3.10- 1.03 = 1.03/v(1-v^2/c^2)

2.07 = 1.03/v(1-v^2/c^2)

v(1-v^2/c^2) = 0.49

squaring both sides

(1-v^2/c^2) = 0.24

0.76 = v^2/c^2

v^2 = 0.76 × (3×10^8)^2

v = 0.873×10^8

v = 2.61×10^8 m/s

s = vt

t = 3.10 years = 3.10×365×24×3600 seconds = 9.7×10^8

s = 2.61×10^8× 9.7×10^8

s = 25.31×10^16 m

5 0
2 years ago
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