SI Prefix Meaning

For any planet, if the planet were to shrink, but have constant mass, what would happen to the value of gravity acceleration (g)

choli [55]

The value of the gravity acceleration on the planet's surface increases

Explanation:

The gravitational acceleration on the surface of a planet is given by:

$g=\frac{GM}{R^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

In this problem, we are told that the planet shrinks, therefore the new radius is smaller than the original radius:

$R'$

while the mass remains the same:

$M'=M$

Therefore, the new acceleration of gravity is

$g'=\frac{GM}{R'^2}$

We see that the value of g is inversely proportional to the square of the radius of the planet: therefore, since $R'$ , it means that $g'>g$ , so when the planet shrinks, the value of the gravity acceleration on the planet's surface increases.

Learn more about gravity:

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5 0

3 years ago

•A radioactive material A (decay constant λA) decays into a material B (decay constant λB) and then into material C (decay const

KiRa [710]

Answer:

The amount of C remaining after time t is

$N_C__{R}} =N_D = (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{-\lambda_A t})e^{-\lambda_B } [e^{-\lambda_C } ]$

Explanation:

We can represent the decay sequence as

$A \to B \to C \to D$

The reason we added D is because we are told from the question that C is also radioactive so it has the tendency to decay

Generally for every decay the remaining radioactive element can be obtained as

$N = N_0 -N_0 e^{- \lambda t}$

Where N is the amount of the remaining radioactive material

$N_0$ is the original amount amount of the radioactive material before decay

and $\lambda$ is the decay constant

Now for the decay from $A \to B$ amount of radioactive element B formed from A after time t can be obtained as

$N_b = N_0 -N_0 e^{- \lambda_A t}$

Where $\lambda _A$ is the decay constant of A

Now for the decay from $B \to C$ amount of radioactive element C formed from A after time t can be obtained as

$N_c = (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{\lambda_A t})e^{-\lambda_B t}$

Where $\lambda _B$ is the decay constant of B

Now for the decay from $C \to D$ amount of radioactive element D formed from A after time t can be obtained as

$N_C__{R}} =N_D = (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{-\lambda_A t})e^{-\lambda_B } [e^{-\lambda_C } ]$

So this amount of D is the reaming amount of the radioactive material C

6 0

2 years ago

At a certain instant a rigid wheel is spinning about its center of mass with angular velocity of magnitude ω and angular acceler

geniusboy [140]

Answer:

a = r√(w⁴ + (alpha)²)

Explanation:

let w be the magnitude of angular velocity

r = radius

v = velocity = wr

angular acceleration, ā = v²/r = (wr)²/r = w²r

Also,

â = dv/dt = rdw/dt

given : alpha = dw/dt

hence, â = (alpha)r

the resultant acceleration, a, by Pythagoras is given as

a = √( â² + ā²) = √(r²w⁴ + r²(alpha)²)

simplifying,

a = √r²(w⁴ + (alpha)²)

a = r√(w⁴ + (alpha)²)

7 0

3 years ago

LAB 11: MASSES AND SPRINGS Part A Theory Please study spring force and oscillation concepts to answer the following questions. 1

Marat540 [252]

Simple harmonic motion is defined as a periodic motion of a point along a straight line, such that its acceleration is always towards a fixed point in that line and is proportional to its distance from that point.

Why is it called simple harmonic motion?

The motion of a particle moving along a straight line with an acceleration whose direction is always towards a fixed point on the line and whose magnitude is proportional to the distance from the fixed point is called simple harmonic motion.

Why is SHM important?

Whilst simple harmonic motion is a simplification, it is still a very good approximation. Simple harmonic motion is important in research to model oscillations for example in wind turbines and vibrations in car suspensions.

What are two basic characteristics of SHM?

Solution : The two basic characteristics of a simple harmonic motion : (i) Acceleration is directly proportional to displacement.

(ii) The direction of acceleration is always towards the mean position, that is opposite to displacement.

Learn more about simple harmonic motion:

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7 0

1 year ago

A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

blsea [12.9K]

Answer:

a) Q = C*emf

b) Reduction in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are connected in series with the battery

Using Kirchoff's voltage law, sum of all voltages in the circuit is zero

Let $V_{R}$ = Voltage dropped across the Resistor

$V_{c}$ = Voltage dropped across the capacitor

Applying KVL;

$emf - V_{R} - V_{c} = 0\\$ .........................(1)

Since the connection is in series, the same current flow through the circuit

$V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C$

Putting $V_{c}$ and $V_{R}$ into equation (1)

$emf - IR - Q/C = 0$

At the final charge, the capacitor in fully charged, and current drops to zero due to equilibrium

$I = 0A\\emf = Q/C\\Q = C* emf$

b) Current starts running through the plate because as the sheet of plastic is inserted between the plates both the electric field intensity and the electric potential reduces. The charge also reduces, then current flows

c) The current through the resistor is the current through the entire circuit ( series connection)

$I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\$

Putting the values of t and I₀ into the formula for I written above

$I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}$

d) NB: The initial charge on the capacitor = C * emf

The final charge will be:

$Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf$

4 0

3 years ago