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LekaFEV [45]
3 years ago
14

What would be an example of higher concentration for a liquid?

Physics
2 answers:
zmey [24]3 years ago
7 0

Answer:

its A

Explanation:

STatiana [176]3 years ago
4 0
I think a, im not a 100% sure tho!!!
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A 40 kg boy is moving upwards upwards in a lift with an acceleration of 2 m /s ^2 what would be the weight felt by him if measuu
neonofarm [45]

Answer:

<h2> 48kg</h2>

f = ma \\

w \:  =  \frac{f}{g}  \\

Explanation:

f \:  = ma \\ f - mg = ma \\ f = ma + mg \\ f = 40 \times 2 + 40 \times 10 \\ f = 480

w =  \frac{f}{g}  \\w =   \frac{480}{10}  \\ w = 48kg

5 0
3 years ago
Why does the closed top of a convertible bulge when the car is riding along a highway? Why does the closed top of a convertible
iren2701 [21]

Answer:

Option D

The air pressure inside the car is greater than the pressure outside.

Explanation:

When considering airflow over and around a surface, from Bernoulli's equation, air flow regions with higher velocity have a lower pressure, and regions with lower velocity have a higher pressure.

The air outside the convertible is moving faster than the air inside the convertible. This leads to a higher pressure zone just below the surface of the roof (inside the car) causing the roof of the convertible to bulge upwards

4 0
3 years ago
What is the difference between: first moment of area and second moment of area?
zhannawk [14.2K]
To determine the centroid of the object first moment of area is used.

To predict the resistance of a shape to bending and deflection which are directly proportional, second moment of area is used.

6 0
3 years ago
I could really use some help on this question guys! Will give brainliest!
aev [14]

Answer:

I think it’s the third one

4 0
2 years ago
Read 2 more answers
A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between
Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

a = \dfrac{F}{m}

Where, F = frictional force

m = mass

Put the value into the formula

a=\dfrac{\mu mg}{m}

a=\mu g

a=0.229\times9.8

a=2.244\ m/s^2

We need to calculate the speed of the sled

Using equation of motion

v^2=u^2-2as

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

v ^2=(5.49)^2-2\times2.244\times3.89

v=\sqrt{(5.49)^2-2\times2.244\times3.89}

v=3.56\ m/s

Hence, The speed of the sled is 3.56 m/s.

8 0
3 years ago
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