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3 years ago

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If the half-life of a 2.0 gram sample of a radionuclide is 15 hours, then the half-life of a 1.0 gram sample of the same radionu

clide would be

1 answer:

Lerok [7]3 years ago

5 0

7.5 hours or 450 minutes. 15/2=7.5

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2 76 Consider a household that uses 14,000 kWh of elec- (tricity per year and 3400 L of fuel oil during a heating sea- son. The

NNADVOKAT [17]

Answer:

The total amount of CO₂ produced will be = 20680 kg/year

The reduction in the amount of CO₂ emissions by that household per year = 3102 kg/year

Explanation:

Given:

Power used by household = 14000 kWh

Fuel oil used = 3400 L

CO₂ produced of fuel oil = 3.2 kg/L

CO₂ produced of electricity = 0.70 kg/kWh

Now, the total amount of CO₂ produced will be = (14000 kWh × 0.70 kg/kWh) + (3400 L × 3.2 kg/L)

⇒ The total amount of CO₂ produced will be = 9800 + 10880 = 20680 kg/year

Now,

if the usage of electricity and fuel oil is reduced by 15%, the reduction in the amount of the CO₂ emission will be = 0.15 × 20680 kg/year = 3102 kg/year

6 0

4 years ago

What is the magnitude of the displacement of a car with an acceleration of 3.75 m/s2 as it increases its speed from 5.20m/s to 1

NeX [460]

Answer:

20.7 m

Explanation:

using v^2=u^2+2as

s=(v^2-u^2)/2a

v=13.5

u=5.2

a=3.75

7 0

3 years ago

A hot air balloon traveled 3 hours at a speed of of 450 km/hr. What distance did it travel?

erastova [34]

Answer:

it traveled 1350 kilometers or 839 miles

Explanation:

You just have to do 450 times three because its 3 HOURS and 450 kilometers per HOUR.

3 0

3 years ago

If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?

Sauron [17]

The kinetic energy of a point body is defined as the product of (1/2) * (m) * (V ^ 2). Therefore, to find the kinetic energy of a body of mass m = 44Kg and velocity V = 10m / s the operation must be done (1/2) * (44) * (10 ^ 2) which results in 2,200 Joules.

5 0

3 years ago

A rock is thrown upward from level ground in such a way that the maximumheight of its flight is equal to its horizontal rangeR.

Snowcat [4.5K]

A) The vertical component of velocity v is taking the rock to a height

Vertical component = $vsin\theta$
The time taken to reach maximum height = $\frac{vsin\theta}{g}$
So total time of rocks flight = $\frac{2vsin\theta}{g}$
Range of rock is due to the horizontal component of velocity = $vcos\theta$
Range = $\frac{2*v*sin\theta*v*cos\theta}{g}$ = $\frac{2*v^2*sin\theta*cos\theta}{g}$
Maximum height = $\frac{g*t^2}{2}$ = $\frac{v^2*sin^2\theta}{2*g}$
Since range = maximum height
We have $\frac{2*v^2sin\theta*cos\theta}{g} = \frac{v^2*sin^2\theta}{2*g}$
$tan\theta = 4$
$\theta = 75.96^0$
So when angle of projection is $\theta = 75.96^0$ range is equal to maximum height reached.
b) We have range = $\frac{2*v^2*sin\theta*cos\theta}{g}$ = $\frac{2*v^2*sin2\theta}{g}$
Maximum of range is reached when $\theta = 45^0$
Maximum range = $\frac{2*v^2}{g}$
c) For range to be equal to maximum height only condition is $tan\theta = 4$ , it does not depend upon acceleration due to gravity and velocity. That angle is a constant.

5 0

3 years ago