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AlexFokin [52]
3 years ago
11

You have a piece of cork with a volume of 2 cm^3 and a density of 210kg/m^3. You hold it under water and release it.

Physics
1 answer:
insens350 [35]3 years ago
7 0

I am sorry if it didn't helped

answers;

Calculate the buoyant force of a piece of cork of 8cm3 that floats in water. Density of cork is 207kg/m3. ?

I need the mass, in order to get the volume to apply t to the Buoyancy formula of: B=(W)object=(m)object(g)

Explanation:

From Archimedes Principle, any object partially or totally submerged in a fluid is buoyed upwards with a force equal to the weight of the displaced fluid.

∴

B

=

ρ

f

l

V

f

l

g

=

1000

k

g

/

m

3

×

8

×

10

−

6

m

3

×

9

,

8

m

/

s

2

=

0

,

0784

N

(assuming the density of water is at standard temperature and pressure, and that the cork is totally submerged as it floats in the water

it's not the answer of your question ⁉️ but it is similar ........

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An arrow strikes a target moving at 75 m/s and embeds itself 15 cm into the target. If the arrow stopped with constant accelerat
zhenek [66]

Answer:

Answer:

4 ms

Explanation:

initial velocity, u = 75 m/s

final velocity, v = 0

distance, s = 15 cm = 0.15 m

Let the acceleration is a and the time taken is t.

Use third equation of motion

v² = u² + 2 a s

0 = 75 x 75 - 2 a x 0.15

a = - 18750 m/s^2

Use first equation of motion

v = u + at

0 = 75 - 18750 x t

t = 4 x 10^-3 s

t = 4 ms

thus, the time taken is 4 ms.

Explanation:

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3 years ago
In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

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  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

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Answer:

Solution:

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