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2 years ago

3. In 1989, Michel Menin of France walked on a tightrope suspended under a

Physics

1 answer:

Tamiku [17]2 years ago

6 0

Answer: 80m

Explanation:

Distance of balloon to the ground is 3150m

Let the distance of Menin's pocket to the ground be x

Let the distance between Menin's pocket to the balloon be y

Hence, x=3150-y------1

Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

40^2= 0^2 + 2 * 10* (3150-y)

1600 = 0 + 63000 - 20y

1600 - 63000 = - 20y

-61400 = - 20y minus cancel out minus on both sides of the equation

61400 = 20y

Hence y = 61400/20

3070m

Hence, recall equation 1

x = 3150 - 3070

80m

I hope this solve the problem.

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Lucille is finding it difficult to play soccer after school. Her doctor thinks that her cells might not be getting enough oxygen

Bad White [126]

Explanation:

Lucile

Lucile cells needs a huge abundance of glucose and oxygen supply in order for her body to function properly and participate in a soccer game.

Playing a soccer game is energy demanding. The body's energy supply is met by the metabolism of glucose in cells using oxygen.

The problem with Lucille's respiratory and circulatory system can make it difficult for her to take part in soccer game because the needed energy would not be available for her to participate in the game.

Respiratory system is saddled with the responsibility of bringing enough oxygen to body system through gaseous exchange and metabolism.
The circulatory system carries the oxygen to the different body cells to where they are need. It also takes the deoxygenated blood to where they can get oxygenated.
We can clearly see that if respiratory and circulatory systems are not functioning well, Lucille's cells will not be able to furnish her with the much needed energy.

Ezra

Ezra needs energy to play basketball well. The energy is produced via the glucose in the starch and the oxygen from breathing.

When starch is broken down, glucose is produced.

How it works;

The digestive, circulatory and respiratory systems are all important here. They ensure that Ezra gets the needed energy from the food he is eating. Proteins also have the capability of producing energy when there is shortage of starch in the body.

Ezra's digestive system through a couple of processes releases glucose from the food he eats especially the bread in the sandwich.

The glucose is stored in the body and made available for use when in need.

The glucose is used to produce energy.

When Ezra breathes, the circulatory system ensures gaseous exchange with the surroundings. Oxygen rich blood is taken the cells where they combine with glucose for their metabolic actions. Oxygen deficient bloods are carried away.

In the cell, a special apparatus called the mitochondria combines glucose with oxygen and energy is liberated. The energy is used by Ezra to run while playing basketball

learn more:

Cellular respiration brainly.com/question/3447259

#learnwithBrainly

7 0

2 years ago

Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot

Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

<u>Explanation:</u>

In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0$

$(m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0$

Taking out ‘mg’ as common and we get

$2 x-50=0$

$2 x=50$

$x=\frac{50}{2}=25 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}$

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$4 m g(x)+(m g)(x-50 c m)=0$

$4 m g x+m g x-50 m g=0$

Taking out ‘mg’ as common and we get

$5 x=50$

$x=\frac{50}{5}=10 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}$

So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

3 0

2 years ago

2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de

kow [346]

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

u = 0

Therefore, the equation becomes

S = 1/2 gt²

Substituting the given values in the above equation

S = 0.5 x 9.8 x 1.5²

= 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

3 0

2 years ago

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

6 0

2 years ago

A projectile is launched horizontally east at a speed of 29.4 M/s towards a wall 88.2 m away. What is the velocity of the projec

Drupady [299]

Time before projectile hits wall

= 88.2 m / 29.4 m/s = 3 seconds

Vertical velocity of projectile after three seconds

= 3*9.8 = 29.4 m/s

Horizontal velocity of projectile after three seconds, assuming no air resistance

= 29.4 m/s (given)

Conclusion:

velocity of projectile when it hits the wall

= < 29.4, -29.4> m/s

= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal

= 41.58 m/s east-bound at 45 degrees below horizontal.

6 0

2 years ago