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2 years ago

Is this correct? (the table)

Physics

1 answer:

Aneli [31]2 years ago

3 0

I’m pretty sure b isn’t right because it’s counting towards multiple strands of DNA.

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When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____

valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

$D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''$

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

$D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26$

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0

3 years ago

What is the common name for tropospheric ozone

Galina-37 [17]

Ozone in troposphere is also know as Bad Ozone, Evil Ozone and Ground Level Ozone.

5 0

2 years ago

How many times is larger than a centigram is a dekagram

S_A_V [24]

A decagram is 1000 times bigger than a centigram

8 0

2 years ago

When a wave strikes a solid barrier it behaves like a basketball hitting a backboard this wave behavior years called?

Mama L [17]

This behavior is called reflection.
Reflection is a change of in direction of the wave when it reaches another medium. Imagine a wave colliding with a glass in a tank of water.
During reflection, some of the initial energy of the wave is lost.
Waves always reflect with at same angle at which it approached the obstacle.

5 0

3 years ago

A ball is dropped from rest at a point 12 m above the ground into a smooth, frictionless chute. The ball exits the chute 2 m abo

Nonamiya [84]

Answer:

29,7 m

Explanation:

We need to devide the problem in two parts:

A) Energy

B) MRUV

<u>Energy:</u>

Since no friction between pint (1) and (2), then the energy conservatets:

Energy = constant ----> Ek(cinética) + Ep(potencial) = constant

Ek1 + Ep1 = Ek2 + Ep2

Ek1 = 0 ; because V1 is zero (the ball is "dropped")

Ep1 = m*g*H1

Ep2= m*g*H2

Then:

Ek2 = m*g*(H1-H2)

By definition of cinetic energy:

m*(V2)²/2 = m*g*(H1-H2) ---> $V2 = \sqrt{(2*g*(H1-H2)}$

Replaced values: V2 = 14,0 m/s

The decomposition of the velocity (V2), gives a for the horizontal component:

V2x = V2*cos(α)

Then the traveled distance is:

X = V2*cos(α)*t.... but what time?

The time what takes the ball hit the ground.

Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²

In the vertical axis:

Y3 = 0 ; Y2 = H2 = 2 m

Reeplacing:

-2 = 14*t + (1/2)*(-9,81)*t²

solving the ecuation, the only positive solution is:

t = 2,99 sec ≈ 3 sec

Then, for the distance:

X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m

6 0

2 years ago

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