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3 years ago

Which important safety measures should be taken during a hurricane? Select two options.

Physics

2 answers:

Roman55 [17]3 years ago

6 0

Answer:

1) move away from the windows.

When abrupt changes in pressure (produced by the hurricane) affect the windows, they can "explode" and be a serious danger for the people.

2) Go to the lowest level of the building.

Is safer than the upper levels because the lower levels suffer less from the strength of the hurricane (being closer to the ground increases the structural resistance), make sure that you have everything you may need in case of an emergency.

max2010maxim [7]3 years ago

6 0

Answer:

The answer is option A and B

Explanation:

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Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.

kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, $4.67\times 10^7$

(b) The number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

Explanation :

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{3.82\text{ days}}$

$k=1.81\times 10^{-1}\text{ days}^{-1}$

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.81\times 10^{-1}\text{ days}^{-1}$

t = time passed by the sample = 12 days

a = initially number of radon atoms = $4.1\times 10^8$

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

$12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}$

$a-x=4.67\times 10^7$

Thus, the number of radon atoms will remain after 12 days is, $4.67\times 10^7$

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = $(4.1\times 10^8)-(4.67\times 10^7)$

The number of radon nuclei have decayed = $3.6\times 10^8$

Thus, the number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

5 0

3 years ago

Does bouncing or sticking produce more impulse?

Elden [556K]

Answer:

Rebounding involves a change in the direction of an object; the before- and after-collision direction is different. ... As mentioned above, if cars rebound upon collision, the momentum change will be larger and so will the impulse. A greater impulse will typically be associated with a bigger force.

<h3>Please mark as brainliest</h3>

4 0

3 years ago

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Based on the graph of velocity over time, which could be the initial velocity and the final velocity for this graph?

dolphi86 [110]

The graph is showing a constant velocity as the line is horizontal. The initial speed is equal to the final speed as there is no changing in acceleration. The second statement is the correct statement

Initial = 2.5 m/s
Final = 2.5 m/s

3 0

3 years ago

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The acceleration of a particle is directly proportional to the square of the time t. When t =0, the particle is at x =24 m. Know

AlexFokin [52]

Answer:

x(t) = ⅟₁₀₈t⁴ + 10t + 24

v(t) = ⅟₂₇t³ + 10

Explanation:

a(t) = C₁t²

velocity is the integral of acceleration

v(t) = ⅓C₁t³ + C₂

position is the integral of velocity

x(t) = (⅟₁₂C₁)t⁴ + C₂t + C₃

x(0) = 24 = (⅟₁₂C₁)0⁴ + C₂0 + C₃

C₃ = 24

x(6) = 96 = (⅟₁₂C₁)6⁴ + C₂6 + 24

72 = 108C₁ + 6C₂

C₂ = 12 - 18C₁

v(6) = 18 = ⅓C₁6³ + C₂

18 = 72C₁ + C₂

18 = 72C₁ + (12 - 18C₁)

6 = 54C₁

C₁ = 1/9

C₂ = 12 - 18(1/9)

C₂ = 10

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3 years ago

In a simple circuit, if you replace one resistor with a resistor of a higher value will the reading the ammeter increase, decrea

Citrus2011 [14]

The ammeter reading (current) will decrease.
V=IR
I = V/R
so I is inversely proportional to R, i.e. increasing R will cause I to decrease

6 0

4 years ago

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