How does natural gas furnes effect the enviroment

a vehicle starts from rsst and after 10 seconds its velocity is 20m/s find the acceleration and the distance travelled.(step by

Naily [24]

Answer:

Acceleration is 2 m/s²

Distance is 100 m

Explanation:

From definition of acceleration, Acceleration is the rate of change in velocity.

• Simplifying the definition, or modifying it;

${ \boxed{ \rm{acceleration = \frac{(final \: velocity - initial \: velocity)}{time} }}}$

• Let's formulate symbols:

${ \boxed{ \rm{a = \frac{v - u}{t} }}}$

a is acceleration
v is final velocity, v = 20 m/s
u is initial velocity, u = 0 m/s [ at rest ]
t is time, t = 10 seconds

${ \rm{a = \frac{20 - 0}{10} = \frac{20}{10} }} \\ \\{ \boxed{ \rm{ \: a = 2 \: {ms}^{ - 2} }}}$

Distance = ut + ½at²

Distance = (0 × 10) + (½ × 2 × 10²)

Distance = 0 + 10²

Distance = 100 meters

8 0

3 years ago

A bike race against the clock takes place on a straight road. Yan drives at 37 km / h and he starts the course 30s before Christ

joja [24]

Given data:

Yan speed;

$u_1=37\text{ km/h}$

Christopher speed;

$u_2=38.9\text{ km/h}$

Christophe starts 30 s later than Yan. Therefore, Christophe takes 30 s less than Yan to reach the same distance.

Part (A)

The distance is given as,

$d=ut$

Let both Yan and Christophe meet at d distance from the start position. Therefore,

$u_1t=u_2(t-30)$

Substituting all known values,

$\begin{gathered} (37\text{ km/h})t=(38.9\text{ km/h})\times(t-30) \\ \frac{(37\text{ km/h})}{(38.9\text{ km/h})}=\frac{(t-30)}{t} \\ 0.95=1-\frac{30}{t} \\ \frac{30}{t}=1-0.95 \\ \frac{30}{t}=0.05 \\ t=\frac{30}{0.05} \\ t=600\text{ s} \end{gathered}$

Therefore, 600 s after Yan's departure Christophe will join him.

Part (B)

The distance is given as,

$d=u_1t$

Substituting all known values,

$\begin{gathered} d=(37\text{ km/h})\times(600\text{ s}) \\ =(37\text{ km/h})\times(600\text{ s})\times(\frac{1\text{ hr}}{3600\text{ s}}) \\ \approx6.17\text{ km} \end{gathered}$

Therefore, Christophe joins Yan after 6.17 km from the start.

3 0

1 year ago

What is the function of eye lens of the human eye

Artyom0805 [142]

Answer:

Lens of the human Eye is a important and one of that most complex sense organ.

Explanation:

Lens of the human eye it helps that in visualizing light and color perception and objects , glance on the human eye structure and function.

Sense organs are much pretty similar to the camera they help us see the objects clear.

A human eye is the 2.3 cm in diameter and all filled some fluid, and there are following parts in eye:- cornea, Retina , Lens ,Pupil , optic nerves.

cornea :- the cornea is the first transparent part of is called cornea, enters the light through the cornea.

Retina :-it is light sensitive layer that consists of nerve cells,then transmitted to the brain through nerves.

Lens:- behind the pupil there is a transparent structure called lens,it shape focus light on the retina.

Pupil:-it control the value of light that enters the human eye.

Optic nerves is are the two types :- (1) cones (2) Rods .

7 0

3 years ago

*8–52. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the to

denis23 [38]

Answer:

μ₁ = 0.1048

μ₂ = 0.1375

Explanation:

Using static equation can find in both point the moment and the forces so:

∑ M = F *d , ∑ F = 0

∑ M A = 0

N₁ * 3 - 200 * 9.81 * 1.5 = 0

N₁ = 981

∑ M y = 0

N₂ + 300 * ³/₅ - 981 - 20 * 9.81 = 0

N₂ = 997.2 N

∑ M C = 0

F₁ * 1.75 - 300 * ⁴/₅ * 0.75 = 0

F₁ = 102.86

∑ M B = 0

300 * ⁴/₅ * 1 - F₂ * 1.75 = 0

F₂ = 137.14 N

The Force F1 and F2 related the coefficients of static friction

F₁ = μ₁ * N₁ ⇒ 102.86 N = μ₁ * 981 ⇒ μ₁ = 0.1048

F₂= μ₂ * N₂ ⇒ 137.14 N = μ₂ * 997 ⇒ μ₂ = 0.1375

8 0

4 years ago

The student throws one cannonball directly upward at 5.0 m/s and simultaneously throws the other cannonball directly downward at

Mademuasel [1]

Let the cannonball be thrown at a height of h above ground.
Then the potential energy of the ball is
V = m*g*h
where
m = the mass of the ball
g = 9.8 m/s²

Also, the kinetic energy of the ball is
K = (1/2)mu²
where
u = 5 m/s, the vertical launch velocity.
Ignore wind resistance.

Because the total energy is preserved, the total energy (n the form of only kinetic energy) when the ball strikes the ground is
(1/2)mV²
where V = vertical velocity when the ball strikes the ground.

Expressions for both the initial and final energy are equal regardless of whether the ball s thrown downward or upward.
Therefore there is no difference in the landing speed.

Answer: There is no difference.

8 0

4 years ago