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3 years ago

Calculate the kinetic energy in joules of a 1,500 kg car that is moving at a speed of 42 km/h

1 answer:

4 0

Data:
KE (Kinetic Energy) = ? (Joule)
m (mass) = 1500 Kg
v (speed) = 42 Km/h
converting to m/s (42 / 3.6), we have: v (speed) = 11.6 m/s

Formula:
$K_{E} = \frac{1}{2} m*v^2$

Solving:
$K_{E} = \frac{1}{2} m*v^2$
$K_{E} = \frac{1}{2} *1500*(11.6)^2$
$K_{E} = \frac{1}{2} *1500*134.56$
$K_{E} = \frac{201840}{2}$
$\boxed{\boxed{K_{E} = 100920\:Joule}}\end{array}}\qquad\quad\checkmark$

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a hot piece of copper is placed in an insulated cup. what is the final temperature of the water and copper?

slavikrds [6]

Answer:

Option C. 30°C.

Explanation:

The following data were obtained from the question:

Mass of water (Mw) = 0.5 Kg

Specific heat capacity of water (Cw) = 4.18 KJ/Kg°C

Initial temperature of water (Tw1) =

22°C

Change in temperature (ΔT) = T2 – Tw1 = T2 – 22°C

Mass of copper (Mc) = 0.5 Kg

Specific heat capacity of copper (Cc) = 0.386 KJ/kg°C

Initial temperature of copper (Tc1) = 115°C

Change in temperature (ΔT) = T2 – Tc1 = T2 – 115°C

Final temperature (T2) =..?

Note: Both the water and the piece copper will have the same final temperature and the heat will be zero since the water will cool the piece of copper.

Thus, we can determine the final temperature of the water and copper as follow:

Q = MwCwΔT + McCcΔT

0 = 0.5 x 4.18 x (T2 – 22) + 0.5 x 0.386 x (T2 – 115)

0 = 2.09 (T2 – 22) + 0.193 (T2 – 115)

0 = 2.09T2 – 45.98 + 0.193T2 – 22.195

Collect like terms

2.09T2 + 0.193T2 = 45.98 + 22.195

2.283T2 = 68.175

Divide both side by the coefficient of T2 i.e 2.283

T2 = 68.175/2.283

T2 = 29.8 ≈ 30°C

Therefore, the final temperature of water and copper is 30°C.

8 0

3 years ago

The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as , where a

Misha Larkins [42]

Answer:

3 times louder

Explanation:

The Loudness in decibel Db L = 10㏒(I/I₀) where I = sound intensity level and I₀ = threshold of hearing = 10⁻¹² W/m².

Now, for Jessica, I₁ = sound intensity level of Jessica's music = 10⁻⁹

and I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₁ = 10㏒(I₁/I₀)

L₁ = 10㏒(10⁻⁹/10⁻¹²)

L₁ = 10㏒(10³)

L₁ = 3 × 10㏒10

L₁ = 30㏒10

L₁ = 30 dB

Now, for Braylee, I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₂ = 10㏒(I₁/I₀)

L₂ = 10㏒(10⁻³/10⁻¹²)

L₂ = 10㏒(10⁹)

L₂ = 9 × 10㏒10

L₂ =90㏒10

L₂ = 90 dB

So, the number of times Braylee's music is louder than Jessica's music is L₂/L₁ = 90 dB/30 dB = 3

So, Braylee's music is 3 times louder than Jessica's music

6 0

3 years ago

What kind of model is shown below?

Rudiy27

D. a foot model

btw this is a joke right cuz there ain’t no picture lol

7 0

3 years ago

What is the middle layer of earth called

irinina [24]

The middle or centre of the Earth is the core. However the middle of the layers from the surface to the centre of the Earth is known as mantle.

6 0

3 years ago

Read 2 more answers

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t

Cloud [144]

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

$W = U_f - U_i$

we know that

$U_f = -\frac{GMm}{6r}$

$U_i = -\frac{GMm}{r}$

now from above formula

$W = -\frac{GMm}{6r} + \frac{GMm}{r}$

$W = \frac{5GMm}{6r}$

so above is the work done to move the mass from surface to given altitude

7 0

3 years ago