The correct answer for the question that is being presented above is this one: "a. only from an instructor or supervisor." Ideally, rewards should be given immediately and frequently but <span>only from an instructor or supervisor to show authority. </span>

3 0

2 years ago

aluminum has a density of 2.7 g/cm3. if a chuck of aluminum with a mass of 16 g is placed in a graduated cylinder partially fill

sweet-ann [11.9K]

When placing the piece of aluminium in water, the level of water will rise by an amount equal to the volume of the piece of aluminum.
Therefore, we need to find the volume of that piece.

Density can be calculated using the following rule:
Density = mass / volume
Therefore:
volume = mass / density
we are given that:
the density = 2.7 g / cm^3
the mass = 16 grams
Substitute in the equation to get the volume of the piece of aluminum as follows:
volume = 16 / 2.7 = 5.9259 cm^3

Since the water level will rise to an amount equal to the volume of aluminum, therefore, the water level will rise by 5.9259 cm^3

5 0

3 years ago

A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis t

Likurg_2 [28]

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

$K = \frac{1}{2}I\omega ^{2}$

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

6 0

2 years ago

Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10

Nezavi [6.7K]

Answer:

t = 23.9nS

Explanation:

given :

Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m

distance d= 1mm=> 0.001

resistor R= 975 ohm

Capacitance can be calculated through the following formula,

C = (ε0 x A )/d

C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001

C = 17.7 x 10^-12 (pico 'p' = 10^-12)

C = 17.7pF

the voltage between two plates is related to time, There we use the following formula of the final voltage

Vc = Vx (1-e^-(t/CR))

75 = 100 x (1-e^-(t/CR))

75/100 = (1-e^-(t/CR))

.75 = (1-e^-(t/CR))

.75 -1 = -e^-(t/CR)

-0.25 = -e^-(t/CR) --->(cancelling out the negative sign)

e^-(t/CR) = 0.25

in order to remove the exponent, take logs on both sides

-t/CR = ln (0.25)

t/CR = -ln(0.25)

t = -CR x ln (0.25)

t = -(17.7 x 10^-12 x 975) x (-1.38629)

t = 23.9 x $10^{-9$

t = 23.9ns

Thus, it took 23.9ns for the potential difference between the deflection plates to reach 75 volts

6 0

3 years ago

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