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3 years ago

Acellus

Physics

1 answer:

rusak2 [61]3 years ago

7 0

Answer:

1.17m

Explanation:

The formula to find distance is d=vt/2

This problem is asking for how far the reflecting object has moved so you need to find the distance from the motion sensor at both times.

(343)(0.115) / (2) = 1.97

(343)(0.0183) /(2) =3.14

After that, all you have to do is find the difference so

3.14 - 1.97

= 1.17

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A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d

oksano4ka [1.4K]

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

$15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm$

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

5 0

2 years ago

3.5 g of a hydrocarbon fuel is burned in a vessel that contains 250. grams of water initially at 25.00 C. After the combustion,

kherson [118]

Answer:

3.) 463 J/g

Explanation:

Heat given by the fuel is used to increase the temperature of the water

so it is given as

$Q = ms\Delta T$

here we will have

m = 250 g

$\Delta T = 26.55 - 25$

now we also know that

$s = 4186$

so we have

$Q = (0.250)(4186)(26.55 - 25)$

$Q = 1622.075 J$

so 3.5 g fuel gives above energy

so per gram of fuel will have total energy given as

$Q = \frac{1622.075}{3.5}$

$Q = 463.45 J/g$

6 0

3 years ago

A 2,300-kg truck is traveling down a highway at 32 m/s. What is the kinetic energy of the truck?

kondor19780726 [428]

m = mass of the truck = 23 00 kg

v = speed of the truck down the highway = 32 m/s

K = kinetic energy of the truck = ?

kinetic energy of the truck down the highway is given as

K = (0.5) m v²

inserting the values

K = (0.5) (2300) (32)²

K = (0.5) (2300) (1024)

K = (1150) (1024)

K = 1177600 J

hence the kinetic energy of the truck comes out to be 1177600 J

6 0

2 years ago

The pulley system below uses a gasoline engine to raise a drill head up through a smooth drill pipe. The engine provides a const

polet [3.4K]

NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question

Answer:

Power = 2702.56 W

Explanation:

Let the power consumed be P

Energy expended = E = mgh

height, h = 5 m

E = 80 * 9.8 * 5

E = 3920 J

$Power = \frac{Energy}{time}$

To calculate the time, t

From F = ma

F = 900 N

900 = 80 a

a = 900/80

a = 11.25 m/s²

From the equation of motion, $s = ut + 0.5at^{2}$

The drill head starts from rest, u = 0 m/s

$5 = 0 * t + (0.5*11.25*t^{2} )\\5 = 5.625t^{2}\\t^{2} = 5/5.625\\t^{2} = 0.889\\t = 0.943 s$

Power, P = E/t

P = 3920/0.0.943

P = 4157.79 W

But Efficiency, E = 0.65

P = 0.65 * 4157.79

Power = 2702.56 W

6 0

2 years ago

Convert the following quantities <br><img src="https://tex.z-dn.net/?f=25m%20%7B%7D%5E%7B2%7D%20%20%5C%3A%20into%20%5C%3A%20cm%2

Bad White [126]

Answer:

$\rm 250000 \; cm^2$

Explanation:

Refer to the attachment.

8 0

2 years ago