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3 years ago

PLZZZZ HELPPPPPP MEEEEEE!!!!!! ASAP!!! ILL GIVE 20 POINTS AND BRAINLIEST!!!

Physics

1 answer:

Snowcat [4.5K]3 years ago

3 0

Answer:

the answers the correct one is C

Explanation:

In radioactive decay processes, a very important concept is the half-life time, which is the time it takes for the parent nuclei to decompose into child nuclei.

expressed this mathematically is

N = N₀ $e^{- \lambda t}$

for the average life time

N₀ / 2 = N₀ $e^{- \lambda T_{1/2}$

T = $\frac{ln 2x}{ \lambda}$

Analyzing this expression, the nuclei that decay are the so-called parents and the new nuclei called children appear.

when checking the answers the correct one is C

You might be interested in

You wish to buy a motor that will be used to lift a 10-kg bundle of shingles from the ground to the roof of a house. The shingle

gogolik [260]

Answer:

$\tau=19.21\ N-m$

Explanation:

It is given that,

Mass of bundle of shingles, m = 10 kg

Upward acceleration of the shingles, $a=1.5\ m/s^2$

The radius of the motor of the pulley, r = 0.17 m

Let T is the tension acting on the shingles when it is lifted up. It can be calculated as :

$T-mg=ma$

$T=m(g+a)$

$T=10\times (9.8+1.5)$

T = 113 N

Let $\tau$ is the minimum torque that the motor must be able to provide. It is given by :

$\tau=r\times T$

$\tau=0.17\times 113$

$\tau=19.21\ N-m$

So, the minimum value of torque is 19.21 N-m. Hence, this is the required solution.

4 0

3 years ago

During a very quick stop, a car decelerates at 28.4 rad/s?. Assume the tires initially rotated in the positive direction and rad

Damm [24]

Answer:

a) 24

b) 3.3 sec

c) 29.8 m/s

d) 48.85 m

Explanation:

α = angular acceleration = - 28.4 rad/s²

r = radius of the tire = 0.32 m

w₀ = initial angular velocity = 93 rad/s

w = final angular velocity = 0 rad/s

θ = angular displacement

Using the equation

w² = w₀² + 2αθ

0² = 93² + 2 (- 28.4) θ

θ = 152.3 rad

n = number of revolutions

Number of revolutions are given as

$n = \frac{\theta }{2\pi }$

$n = \frac{152.3 }{2(3.14) }$

$n = 24$

t = time taken to stop

using the equation

w = w₀ + αt

0 = 93 + (- 28.4) t

t = 3.3 sec

v₀ = initial velocity of the car

initial velocity of the car is given as

v₀ = r w₀ = (0.32) (93) = 29.8 m/s

v = final velocity = 0 m/s

a = linear acceleration = rα = (0.32) (- 28.4) = - 9.09 m/s²

d = distance traveled by car before stopping

Using the equation

v² = v₀² + 2 a d

0² = 29.8² + 2 (- 9.09) d

d = 48.85 m

8 0

4 years ago

When this current is closed which way does the current flow

Anastaziya [24]

Well, Godess, that's not a simple question, and it doesn't have
a simple answer.

When the switch is closed . . .

"Conventional current" flows out of the ' + ' of the battery, through R₁ ,
then through R₂ , then through R₃ . It piles up on the right-hand side of
the capacitor (C). It repels the ' + ' charges on the left side of 'C', and
those flow into the ' - ' side of the battery. So the flow of current through
this series circuit is completely clockwise, around toward the right.

That's the way the first experimenters pictured it, that's the way we still
handle it on paper, and that's the way our ammeters display it.

BUT . . .

About 100 years after we thought that we completely understand electricity,
we discovered that the little tiny things that really move through a wire, and
really carry the electric charge, are the electrons, and they carry NEGATIVE
charge. This turned our whole picture upside down.

But we never changed the picture ! We still do all of our work in terms of
'conventional current'. But the PHYSICAL current ... the actual motion of
charge in the wire ... is all exactly the other way around.

In your drawing ... When the switch is closed, electrons flow out of the
' - ' terminal on the bottom of the battery, and pile up on the left plate of
the 'C'. They repel electrons off of the right-side of 'C', and those then
flow through R₃ , then through R₂ , then through R₁ , and finally into the
' + ' terminal on top of the battery.

Those are the directions of 'conventional' current and 'physical' current
in all circuits.

In the circuit of YOUR picture that you attached, there's more to the story:

Battery current can't flow through a capacitor. Current flows only until
charges are piled up on the two sides of 'C' facing each other, and then
it stops.

Wait a few seconds after you close the switch in the picture, and there is
no longer any current in the loop.

To be very specific and technical about it . . .

-- The instant you close the switch, the current is

(battery voltage) / (R₁ + R₂ + R₃) amperes

but it immediately starts to decrease.

-- Every (C)/((R₁ + R₂ + R₃) seconds after that, the current is

e⁻¹ = about 36.8 %

less than it was that same amount of time ago.

Now, are you glad you asked ?

4 0

3 years ago

A theme park creates a new kind of water wave pool with large waves caused by constructive interference. There are two wave gene

melomori [17]

Answer:

10.0 m

Explanation:

Since there is no amplitude at the point of the swimmer, we have destructive interference.

So, the path difference ΔL = L₂ - L₁ where L₁ = swimmer's shorter distance from one generator = 9.0 m and L₂ = swimmer's longer distance from the other generator = 14.0 m. ΔL = 14.0 m - 9.0 m = 5.0 m

Also, since we have destructive interference, ΔL = (n + 1/2)λ where n = number of wavelengths and λ = wavelength of waves

For maximum wavelength, n = 0

So, ΔL = (n + 1/2)λ

ΔL = (0 + 1/2)λ

ΔL = λ/2

λ/2 = ΔL

λ = 2ΔL

λ = 2 × 5.0 m

λ = 10.0 m

So, the longest wavelength that will produce this interference pattern is λ = 10.0 m

8 0

3 years ago

A cannonball is launched from the ground with 48279 Joules of kinetic energy. How much gravitational potential energy will the c

Mamont248 [21]

When the cannonball is shot upwards, kinetic energy converts to gravitational potential energy (GPE).

As the ball rises, speed decreases and height increases.

So when the cannonball has reached its maximum height, all of the Kinetic energy has transferred into gravitational potential energy.

(Because at the max height, the cannonball for a brief moment has no velocity, and thus no kinetic energy)

So the GPE is 48279 Joules at it's maximum height.

6 0

4 years ago