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2 years ago

10

A force of 8480 N is applied to a cart or decelerate is at a rate of 32.0 m/s2. What is the mass of the cart?

1 answer:

Otrada [13]2 years ago

4 0

I believe it to be 265kg.

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Use the exact values you enter to make later calculations. You repeat the same experiment that you did in the lab with the force

frutty [35]

Answer:

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3 years ago

WILL GIVE BRAINLIEST!!!

Sergeeva-Olga [200]

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3 years ago

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Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.

DaniilM [7]

Answer: $2(10)^{-9} mg$

Explanation:

We know the total diameter of the cell (assumed spherical) is:

$d=1.9\mu m=1.9(10)^{-6} m$

Then its total radius $r=\frac{d}{2}=\frac{1.9(10)^{-6} m}{2}=9.5(10)^{-7} m$

On the other hand, we know the thickness of the cell wall is $r_{t}=60 nm= 60(10)^{-9} m$ and its density is the same as water ( $\rho=997 kg/m^{3}$ ).

Since density is the relation between the mass $m$ and the volume $V$ :

$\rho=\frac{m}{V}$

The mass is: $m=\rho V$ (1)

Now if we are talking about this cell as a thin spherical shell, its volume will be:

$V=\frac{4}{3}\pi R^{3}$ (2)

Where $R=r-r_{w}=9.5(10)^{-7} m - 60(10)^{-9} m$

Then:

$V=\frac{4}{3}\pi (9.5(10)^{-7} m - 60(10)^{-9} m)^{3}$ (3)

$V=2.952(10)^{-18} m^{3}$ (4)

Substituting (4) in (1):

$m=(997 kg/m^{3})(2.952(10)^{-18} m^{3})$ (5)

$m=2.94(10)^{-15} kg$ (6)

Knowing $1 kg=1000 g$ and $1 mg=0.001 g$ :

$m=2.94(10)^{-15} kg=2(10)^{-9} mg$

7 0

3 years ago

True or False? Mechanical waves do NOT need a medium to travel through.

stellarik [79]

Answer:

False.

Explanation:

Mechanical waves require a medium in order to transport their energy from one location to another.

6 0

2 years ago

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Sixty identical drippers, each with a hole of diameter 1.00 mm, are used to water a yard. If the water in the main pipe of diame

VashaNatasha [74]

Answer:

a

$V = 5.30 *10^{-2} \ m^3$

b

$v_1 = 0.3127 \ m/s$

Explanation:

From the question we are told that

The number of identical drippers is n = 60

The diameter of each hole in each dripper is $d = 1.0 \ mm = 0.001 \ m$

The diameter of the main pipe is $d_m = 2.5 \ cm = 0.025 \ m$

The speed at which the water is flowing is $v = 3.00 \ cm/s = 0.03 \ m/s$

Generally the amount of water used in one hour = 3600 seconds is mathematically represented as

$V = A * v * 3600$

Here A is the area of the main pipe with value

$A = \pi * \frac{d^2}{4}$

=> $A = 3.142 * \frac{0.025^2}{4}$

=> $A = 0.0004909 \ m^2$

So

=> $V = 0.0004909 * 0.03 * 3600$

=> $V = 5.30 *10^{-2} \ m^3$

Generally the area of the drippers is mathematically represented as

$A_1= n * \pi \frac{d^2}{4}$

=> $A_1 = 60 * 3.142 * \frac{0.001 ^2}{4}$

=> $A_1 = 4.713 *10^{-5} \ m^2$

Generally from continuity equation we have that

$Av = A_1 v_1$

=> $0.0004909 * 0.03 = 4.713 *10^{-5} * v_1$

=> $v_1 = 0.3127 \ m/s$

3 0

3 years ago