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Alla [95]
3 years ago
10

Abox has a mass of 14.4 kg. The length is 4 m longthe width is 1 m and the height is 5 m. What is the density of the box?

Physics
1 answer:
valkas [14]3 years ago
3 0

Answer:

0.72 kg per cubic m

Explanation:

Mass = 14.4 kg

Volume = lbh = 4*1*5 = 20 cubic m

\because \: density \\  \\  =  \frac{mass}{volume}  \\  \\  =  \frac{14.4}{20}  \\  \\  = 0.72 \: kg {m}^{ - 3}

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3 years ago
(a) A 1.00-μF capacitor is connected to a 15.0-V battery. How much energy is stored in the capacitor? ________ μJ (b) Had the ca
gulaghasi [49]

Answer:

(a) E_{ c} = 112.5 \mu J

(b) E'_{ c} = 18 \mu J

Solution:

According to the question:

Capacitance, C = 1.00\mu F = 1.00\times 10^{- 6} F

Voltage of the battery, V_{b} = 15.0 V

(a)The Energy stored in the Capacitor is given by:

E_{c} = \frac{1}{2}CV_{b}^{2}

E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}

E_{ c} = 112.5 \mu J

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

E'_{c} = \frac{1}{2}CV'_{b}^{2}

E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}

E'_{ c} = 18 \mu J

6 0
4 years ago
Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of 38.0° with the normal to th
iren [92.7K]

Answer:

n_glass = 1.404

Explanation:

In order to calculate the index of refraction of the light you use the Snell's law, which is given by the following formula:

n_1sin\theta_1=n_2sin\theta_2         (1)

n1: index of refraction of vacuum = 1.00

θ1: angle of the incident light respect to normal of the surface = 38.0°

n2: index of refraction of glass = ?

θ2: angle of the refracted light in the glass respect to normal = 26.0°

You solve the equation (1) for n2 and replace the values of all parameters:

n_2=n_1\frac{sin\theta_1}{sin\theta_2}=(1.00)\frac{sin(38.0\°)}{sin(26.0\°)}\\\\n_2=1.404

The index of refraction of the glass is 1.404

3 0
3 years ago
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