Write any three applicataion pressure in our daily life ?

How does air resistance affect the acceleration of falling objects?

BlackZzzverrR [31]

Air resistance opposes the motion of falling objects. Dense air prefers to stay at low heights. More dense the air is, more air resistance it produces. Hence acceleration of falling objects decreases. Hope this helps!

7 0

2 years ago

Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =

Semenov [28]

For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
$r_1+r_2=L km$
The second equation is going to the total force acting on the charge q3:
$F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}$
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
$0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}$
Let's solve this for r_1^2:
$0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0$
Now we have a quadratic equation with following parameter:
$a=2\\ b=2L\\ c=-L^2$
We know that two solutions are:
$r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\$
We need a positive solution. When we plug in all the numbers we get:
$r_1=0.915\cdot 10^6$km$

6 0

3 years ago

Sam heaves a 16lb shot straight upward, giving it a constant upward acceleration from rest of 35 m/s^2 for 64.0 cm. He releases

makvit [3.9K]

Answer:

6.69 m/s

4.483 m

1.42s

Explanation:

Given that:

Initial Velocity, u = 0

Final velocity, v =?

Acceleration, a = 35m/s²

1.) using the relation :

v² = u² + 2as

v² = 0 + 2(35) * 64*10^-2m

v² = 70 * 0.64

v = sqrt(44.8)

v = 6.693

v = 6.69 m/s

B.) height from the ground, h0 = 2.2

How high ball went , h:

Using :

v² = u² + 2as

Upward motion, g = - ve

0 = 6.69² + 2(-9.8)*(h - 2.2)

0= 6.69² - 19.6(h - 2.2)

44.7561 + 43.12 - 19.6h = 0

19.6h = 44.7561 - 43.12

h = 87.8761 / 19.6

h = 4.483 m

C.)

vt - 0.5gt² = h - h0

6.69t - 0.5(9.8)t²

6.69t - 4.9t² = 1.83 - 2.2

-4.9t² + 6.69t + 0.37 = 0

Using the quadratic equation solver :

Taking the positive root:

1.4185 = 1.42s

5 0

2 years ago

A Yugo can accelerate from rest to a speed of 28 m/s in 20 s. What is the average acceleration of the car? What distance does it

harina [27]

Answer:

Explanation:

a=28/20

a=1.4 m/s²

V²=v0²+2a.D

28²=0+2x1.4D

784 = 2.8D

D = 280 m

6 0

2 years ago

How is mercury barometer constructed ?<br>

olganol [36]

Answer:

A mercury barometer is a device use to measure stomspheric pressure and is constructed as following:

A mercury barometer requires a tube which has one close end, and one open end.
Tube is placed upside down in a beaker in such a way so that one end open in the beaker and the other remain outside of the beaker.

The barometric liquid (mercury) is then filled in the tube by pouring mercury liquid in the beaker.

The position of tube creates vacuum between the closed end of the tube and liquid surface and the Mercury has high density that is why used as the liquid to measure pressure.

4 0

3 years ago

Write any three applicataion pressure in our daily life ?​

Write any three applicataion pressure in our daily life ?