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2 years ago

Help 100 points will give brainleyest

Physics

1 answer:

Airida [17]2 years ago

7 0

Answer:

Explanation:

The sound waves are compromised. It decrompesses each time which makes the sound gets lower and lower.

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I need help asap pls

tigry1 [53]

no.9 indeed all matters are made out of atoms which is smaller than protons and neutrons

6 0

2 years ago

A child looks at his reflection in a spherical Christmas tree ornament 8.0 cm in diameter in season that the image of his face i

malfutka [58]

From the information given,

diameter of ornament = 8

radius = diameter/2 = 8/2

radius of curvature, r = 4

Recall,

focal length, f = radius of curvature/2 = 4/2

f = 2

Recall,

magnification = image d

8 0

1 year ago

In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li

Alenkinab [10]

Answer:

Total energy saving will be 0.8 KWH

Explanation:

We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW

30 bulbs are of power 60 W

So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW

Total power of 80 bulbs = 1.8+5 = 6.8 KW

Total time = 3 hour

We know that energy $E=power\times time=6.8\times 3=20.4KWH$

Now power of each CFL bulb = 25 W

So power of 80 bulbs = 80×25 = 2000 W = 2 KW

Energy of 80 bulbs = 2×3 = 6 KWH

So total energy saving = 6.8-6 = 0.8 KWH

6 0

3 years ago

Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25

Sveta_85 [38]

Answer:

8.6 miles

Explanation:

We need to calculate the components of the total displacement along the east-west and north-south directions first.

In the first part, Erica moves 5.2 miles at 25∘ north of east. So the components of this displacement along the two directions are:

East: $d_{1x} = 5.2 cos 25^{\circ}=4.7 mi$

North: $d_{1y} = 5.2 sin 25^{\circ}=2.2 mi$

In the second part, Erica moves 5.0 miles north. So, the components of this displacement are:

East: $d_{2x}=0$

North: $d_{2y} = 5.0 mi$

So the components of the total displacement are

East: $d_x = d_{1x}+d_{2x}=4.7 + 0 = 4.7 mi$

North: $d_y = d_{1y}+d_{2y}= 2.2 + 5.0 = 7.2 mi$

Therefore the magnitude of the displacement, which is the straight-line distance from the starting point to the end of the race, is

$d=\sqrt{d_x^2 +d_y^2}=\sqrt{4.7^2+7.2^2}=8.6 mi$

5 0

3 years ago

(A) how much work is done when you push a crate horizontally with 100 N across a 10-m factory floor?

Mila [183]

Answer:

A) 1000 joules

Explanation:

In general work is given by the equation:

$W=\intop_{a}^{b}\overrightarrow{F}\cdot d\overrightarrow{s}$ (1)

A) With $\overrightarrow{s}$ the displacement and $\overrightarrow{F}$ the force applied, because the force and the displacement are parallel (the crate is pushed horizontally) $\overrightarrow{F}\cdot d\overrightarrow{s}$ is simply $F\,ds$ , and because the path is a straight line and the force is constant work is:

$W=FS$ (2),

$W=(100)(10)=1000\,J$

B) The work-energy theorem says that the total work on a body is equal to the change on kinetic energy:

$W_{tot}=\varDelta K$ (3)

The total work on the crate is the work done by the push and plus the work of the friction $W_{tot}=W + W_{f}$ (4) , as (A) because forces are parallel to the displacement $W= FS$ (5) and $W_{f}=-fS$ (6), the due friction always has negative sign because is opposite to the displacement, using (6), (5) and (4) on (3):

$FS-fS=\varDelta K$ (3)

$1000-(70)(10)=300\,J$

C) The energy is lost by friction, so the amount of energy turned into heat is the work the friction does:

$Q=fS=(70)(100)=700\,J$ (3)

6 0

3 years ago