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Physics

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mina [271]4 years ago

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Calcium-51 has a half-life of 4.5 days. Only 0.75 gram remains of a sample that was initially 12 grams. How old is the sample of

Over [174]

How many times did the original sample lose 50% of its radioactivity ?

-- Start with. . . . . . . . . . . . 12 grams.
-- Lose half of it once. . . . . . 6 grams left.
-- Lose half of it again . . . . . 3 grams left.
-- Lose half of it again . . . . . 1.5 grams left.
-- Lose half of it again . . . . . 0.75 gram left.

-- How many times did it lose half ? 4 times.

-- How long does it take to lose half ? 4.5 days.
(That's why it's called the 'half-life'.)

-- How long did it take to lose half, 4 times ?

(4 x 4.5 days) = 18 days .

6 0

3 years ago

Read 2 more answers

An ac generator with peak voltage 100 volts is placed across a 10-? resistor. What is the average power dissipated?

Romashka [77]

The correct answer is: Average Power = 500 W

Explanation:
Root-mean square voltage = Vrms = Vpeak /√2 = 100 / √2 volts
Resistance = R = 10 Ω
Average power = Pavg = (Vrms)^2<span> / R </span>= (100 * 100) / (2 * 10) = <span>500 W</span>

3 0

4 years ago

Derive the expression for the range of the ball as a function of its height. To do this you should use the following steps. In t

AVprozaik [17]

Explanation:

Thanks for this....

its very helpful...

4 0

3 years ago

What happens to the electrostatic force between two charged particles if the distance between them is doubled?

Stells [14]

According to Coulomb's Law , The size of the force varies inversely as the square of the distance between the two charges. So ,if the distance between the two charges is doubled, the electrostatic force will become weak by one fourth of the original force.

5 0

3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

4 years ago