6. waves that travel through Earth<br><br><br><br> Help me with the answer please yall‼️

schepotkina [342]

This passage is about how people research food, how much calories, much much energy you get from it, etc. It also explains the different kinds of food get a certain amount of calories. Some calories burn quicker than the other.

5 0

2 years ago

When a pair of balanced forces acts on an object, the net force that results is.

polet [3.4K]

D. Equal to zero.

Because the forces balance each other.

7 0

2 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

2 years ago

A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c

Leya [2.2K]

Answer:

$T=83.37N$

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

$\sum F_y: T-mg=F_c$

Where $F_c$ is the centripetal force and is given by:

$F_c=ma_c=m\frac{v^2}{r}$

Replacing and solving for T:

$T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N$

8 0

3 years ago

I need help it is due today

siniylev [52]

Answer:

Option 3. The tennis ball began from rest and rolls at a rate of 14.7 m/s safer 1.5 seconds.

Explanation:

To know the the correct answer to the question, it is important that we know the definition of acceleration.

Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as:

a = (v – u) /t

Where

a => acceleration

v => final velocity

u => Initial velocity

t => time

With the above information in mind, let us consider the options given in the question above to know which conform to the difinition of acceleration.

For Option 1,

We were told that the tennis ball has the following:

Distance = 4 m

Time = 1.5 s

This talks about the speed and not the acceleration.

Speed = distance / time

For Option 2,

We were only told about the average speed and nothing else.

For Option 3,

We were told that the tennis ball have the following:

Initial velocity (u) = 0 m/s

Final velocity (v) = 14.7 m/s

Time = 1.5 s

This talks about the acceleration.

a = (v – u) /t

For Option 4,

We were only told that the tennis rolls to the right at an average speed. This talks about the average velocity. We need more information like time to justify the acceleration.

From the above illustrations, option 3 gives the correct answer to the question.

8 0

2 years ago

6. waves that travel through Earth Help me with the answer please yall‼️