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andrew-mc [135]

2 years ago

14

which of the following can best help you in a risky situation? stay close to the quieter kids have an excuse ready so you can le

ave hang around but don't participate participate only a little to stay safe

2 answers:

monitta2 years ago

8 0

Answer: Option (b) is the correct answer.

Explanation:

A risky situation is a situation in which there is no idea what the consequence would be as there might be a loss, injury or gain of something. So, basically a risky situation is not predictable.

For example, Ravi didn't attended Juhi's birthday party and when Juhi asked him the reason then Ravi made an excuse that he was out of station at that time due to some office work.

Therefore, we can conclude that out of the given options, have an excuse ready so you can leave best help you in a risky situation.

lesya692 [45]2 years ago

3 0

Have an excuse ready so you can leave

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Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago

A 100 N force is applied to a 500 kg crate resting on frictionless wheels.

Alla [95]

Answer:

<h2>0.2 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{100}{500} = \frac{1}{5} = 0.2 \\$

We have the final answer as

<h3>0.2 m/s²</h3>

Hope this helps you

6 0

2 years ago

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What is <br>M=35g, V=17cm³

cestrela7 [59]

Answer: 25

Explanation:

6 0

2 years ago

You push on a box with 100 N of force, causing it to accelerate at 5 m/s?.

Xelga [282]

<h3><u>Given</u><u>:</u><u>-</u></h3>

Force,F = 100 N

Acceleration,a = 5 m/s²

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the mass of the box .

<h3><u>Formula</u><u> </u><u>used:-</u><u> </u></h3>

$\bf \: Force = Mass \times Acceleration$

<h3><u>Solution:-</u><u> </u></h3>

$\sf \: Force = Mass × Acceleration$

★ Substituting the values in the above formula,we get:

$\sf \implies \: 100 = Mass \times 5$

$\sf \implies \: Mass = \cancel\dfrac{100}{5}$

$\sf \implies \: Mass = 20 \: kg$

4 0

2 years ago

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ASK YOUR TEACHER Red blood cells often can be charged. Two red blood cells are separated by 1.11 m and have an attractive electr

makvit [3.9K]

Answer:

q2 = -1.61*10^-5 C.

Explanation:

It was given that,

F = 0.985N

q1 = +8.40 X10-6 C

q2 = ?

r = 1.11 m

k = 9 x 10^9 (standard)

It generally follows that, if force is attractive, charge will be negative.

force, F = kq1q2/r^2

0.985 = 9*10^9*8.40*10^-6*q2/1.11^2

75600q2 = 0.985*1.11^2

75600q2 = 1.2136

q2 = 1.2136/75600 = 1.60529

q2 = -1.61*10^-5 C.

7 0

2 years ago

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