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2 years ago

14

Sven's team put its insulated ice cube into a sealed plastic container. Which

1 answer:

kaheart [24]2 years ago

6 0

They are trying to prevent convection

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An object is moving with uniform speed in a circle of radius r. Calculate the distance and displacement

dimaraw [331]

Answer and Explanation:

distance will be 2×3.14 (pie)×r

displacement will be 2r (diameter)

the motion is uniform circular motion as the object is moving in a circular path with uniform motion

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2 years ago

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Balanced forces acting on an object cause the object to accelerate. Please select the best answer from the choices provided T F

Pani-rosa [81]

False
Balanced forces mean that there is no net force acting on the object. therefore, the object will not accelerate.

7 0

2 years ago

Sally Sizzle builds a circuit by connecting three resistors in parallel with a 9V battery. The current in this circuit is 3.6 am

Andru [333]

Then answer is C 9 ohms

7 0

3 years ago

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Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle

marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

$n_1sin\theta_1 = n_2sin\theta_2$

Where,

$\theta =$ Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

$n_1 sin\theta_1 = n_2sin\theta_2$

$(1.54) sin\theta_1 = (1.33)sin(90)$

$sin\theta_1 = \frac{1.33}{1.54}$

$\theta = sin^{-1}(\frac{1.33}{1.54})$

$\theta = 59.72\°$

Therefore the $\alpha_{max}$ would be equal to

$\alpha = 90\°-\theta$

$\alpha = 90-59.72$

$\alpha = 30.27\°$

Therefore the largest value of the angle α is 30.27°

3 0

3 years ago

In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the ot

Fittoniya [83]

Answer:

$Q=7.9\times 10^{-10}\ C$

Explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

$A=5\times 10^{-3}\ m^2$

We Capacitance given as

For air

$C_1=\dfrac{\varepsilon _oA}{d}$

$C_1=\dfrac{\varepsilon _oA}{d}$

$C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}$

$C_1=2.2\times 10^{-11}\ F$

$C_2=\dfrac{K\varepsilon _oA}{d}$

$C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F$

$C_2=6.6\times 10^{-11}\ F$

Net capacitance

C=C₁+C₂

$C=8.8\times 10^{-11}\ F$

We know that charge Q given as

Q= C V

$Q=12\times 6.6\times 10^{-11}\ C$

$Q=7.9\times 10^{-10}\ C$

6 0

2 years ago