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swat32
3 years ago
5

in terms of loudness and pitch how would you compare the sounds made by a mosquito flying near your ear and the croaking of a bu

llfrog ?
Physics
1 answer:
77julia77 [94]3 years ago
6 0

The mosquito would have a higher pitch but quiet and the croaking bullfrog would be a lower pitch but loud.

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A 4-kg cat is resting on a bookshelf that is 2 m high. What is the cats gravitational potential energy relative to the floor if
ra1l [238]
The most exact answer is 78.4J also in this kind of options we can say answer "d"
3 0
3 years ago
An aircraft travels a distance of 50km in a straight line
Juliette [100K]

Answer:

200 m/s

Explanation:

v = distance / time = 50km/250s = 50000m/250s = 200 m/s

6 0
2 years ago
Warning triangles, flares, a vehicle's hazard lights, or emergency vehicles ahead, are all clues that you might be approaching _
daser333 [38]

Answer:

B. A collision scene

Explanation:

It could have been a parade ceremony, but, if you notice the vehicle's hazard lights or an emergency vehicle ahead, it is common sense to figure that they is a collision scene nearby.

8 0
3 years ago
A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m
kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

3 0
3 years ago
How much energy (in Joules) is released when 12.0 g of water cools from 20.0 °C to 11.0 °C? This is a grade 10 question from the
KATRIN_1 [288]

Answer: - 452.088joule

Explanation:

Given the following :

Mass of water = 12g

Change in temperature(Dt) = (11 - 20)°C = - 9°C

Specific heats capacity of water(c) = 4.186j/g°C

Q = mcDt

Where Q = quantity of heat

Q = 12g × 4.186j/g°C × - 9°C

Q = - 452.088joule

7 0
3 years ago
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