What to do if carbon monoxide detector is going off

a lorry travels 3600m on a test track accelerating constantly at 3m/s squared from standstill. what is the final velocity (3 sig

suter [353]

The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

$2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}$

Then, the final velocity will be

$v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s$

Thus, the final velocity of the truck is found as 146.969 m/s.

8 0

3 years ago

7. A child of mass m starts from rest and slides without friction from a height h along a curved waterslide (Fig. P5.46). She is

marissa [1.9K]

The mechanical energy of the girl will be conserved because the system is isolated and the initial potential energy will be equal to final kinetic energy.

<h3>What is the law of conservation of energy?</h3>

The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.

The change in the potential energy of the launched from a height into the pool without friction from the given height h is calculated by applying the following kinematic equation.

ΔP.E = ΔK.E

where;

ΔP.E is change in potential energy of the child
ΔK.E is change in the kinetic energy of the child

mghf - mghi = ¹/₂mv² - ¹/₂mu²

where;

m is the mass of the girl
g is acceleration due to gravity
hi is the initial height of the girl
hf is the final height when she is launched into the pool
u is the initial velocity
v is the final velocity of the girl

Thus, for every closed or isolated system such as this case, mechanical energy is always conserved because the initial potential energy of the girl will be converted into her final kinetic energy.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

#SPJ1

4 0

1 year ago

You will be colliding two carts in this lab. Select all the systems for which you predict the total momentum of the system will

baherus [9]

From the momentum conservation we know that the initial momentum is equal to the final momentum. The momentum in a singular way can be defined as the product between the mass and the velocity of an object. In the presented system, however, there are two objects, therefore the mass of both and the speed of both, before and after the collision must be taken into account. Mathematically we could describe this as

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Here,

$m_{1,2}$ = Mass of each object

$u_{1,2}$ = Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

From here we can realize that it is necessary to use the system on both cars to be able to predict what will happen either with their masses, or their speeds.

The correct answer is C.

6 0

3 years ago

A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar

Artist 52 [7]

Answer:

a) v₀ = 32.64 m / s , b) x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

x = v₀ₓ t

y = $v_{oy}$ t - ½ g t²

We look for the components of speed with trigonometry

sin 43 = v_{oy} / v₀

cos 43 = v₀ₓ / v₀

v_{oy} = v₀ sin 43

v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

t = x / v₀ cos 43

y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

y = x tan 43 - ½ g x² / v₀² cos² 43

1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

v₀ = √ (35280 / 33.11)

v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

y = $v_{oy}$ t - ½ g t²

.y = v₀ sin43 t - ½ g t²

25 = 32.64 sin 43 t - ½ 9.8 t²

4.9 t² - 22.26 t + 25 = 0

t² - 4.54 t + 5.10 = 0

We solve the second degree equation

t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

t = (4.54 ± 0.46) / 2

t₁ = 2.50 s

t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

x = v₀ₓ t

x = v₀ cos 43 t

x = 32.64 cos 43 2.50

x = 59.68 m

8 0

3 years ago

9. Pick the best example of Newton's Second Law in action.

madreJ [45]

Answer:

A rocket taking off from earth which pushes gasses in one direction and the rocket in

the other

7 0

2 years ago