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1 year ago

7

If a sample of a radioactive isotope has a half-life of 1 year, how much of the original sample will be left at the end of the s

econd year?

1 answer:

Tasya [4]1 year ago

4 0

Answer:

1/4 of the original

Explanation:

That would be TWO half lives:

1/2 * 1/2 = 1/4 <======= 1/4 would be left

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5. Describe the shape of the waveform in the secondary coil for a sine, square and triangle wave in the primary coil. How does t

Volgvan

Answer:

When primary coil is exited by sin wave,this will result in sin wave in secondary coil as well.According to law,flux induced in the secondary coil will have same waveform as in the primary coil.

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3 years ago

Suppose you increase your walking speed from 7 m/s to 15 m/s in a period of 2 m. What is your acceleration?

likoan [24]

Acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed) = (15 m/s - 7 m/s) = 8 m/s

time for the change = 2 minutes = 120 seconds

Acceleration = (8 m/s) / (120 seconds)

Acceleration = 0.067 m/s²

7 0

3 years ago

An ice cream maker has a refrigeration unit which can remove heat at 120 Js'. Liquid ice

Rom4ik [11]

Answer:

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C is 45,360 J

Explanation:

The given parameters for the refrigeration unit and the ice cream are;

The power of the refrigeration unit = 120 J/s

The mass of the liquid ice cream, m = 0.6 kg

The initial temperature of the liquid ice cream, T₁ = 20°C

The freezing point temperature of the ice cream, T₂ = -16°C

The specific heat capacity of the ice cream, c = 2,100 J/kg⁻¹·°C⁻¹

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, ΔQ, is given as follows;

ΔQ = m × c × ΔT

Where;

ΔT = T₁ - T₂

∴ ΔQ = m × c × (T₁ - T₂)

Therefore, by substituting the known values, we have;

ΔQ = 0.6 × 2,100 × (20 - (-16)) = 45,360

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C = ΔQ = 45,360 J.

8 0

3 years ago

A 4.0-n puck is traveling at 3.0 m/s. it strikes an 8.0-n puck, which is stationary. the two pucks stick together. their common

Natalka [10]

<span>1.0 m/s Momentum = mass x velocity Total Momentum before any collision = total momentum afterwards 4.0 x 3.0= 12 :g x momentum before (x g because using weight) Afterwards, if the velocity of the two joined is v then we get: 'momentum x g'=12v so 12v=12 so v=1m/s</span>

3 0

3 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

$F_D = c_D A \frac{\rho V^2}{2}$

Where,

F_D = Drag Force

$c_D$ = Drag coefficient

A = Area

$\rho$ = Density

V = Velocity

Our values are given by,

$c_D = 0.5$ (That is proper of a cone-shape)

$A = 9m^2$

$\rho = 1.2Kg/m^3$

$V = 6.5m/s$

Part A ) Replacing our values,

$F_D = 0.5*9*\frac{1.2*6.5^2}{2}$

$F_D = 114.075N$

Part B ) To find the torque we apply the equation as follow,

$\tau = F*d$

$\tau = (114.075N)(7)$

$\tau = 798.525N.m$

3 0

3 years ago