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Katena32 [7]
3 years ago
10

The word “virtual” refers to something that exists in effect but not in actual fact. How does this definition relate to the virt

ual image you see of yourself in a plane mirror?
Physics
2 answers:
atroni [7]3 years ago
8 0
Because you see yourself the opposite way in a mirror. So yes your “seeing” yourself but not how everyone else sees you.
givi [52]3 years ago
5 0

Answer:

The image seems to be behind the mirror, but nothing is really there.

Explanation:

Sample response edge 2021

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A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

8 0
2 years ago
Question 2 of 10
Mekhanik [1.2K]

the awnser to ur question is D

6 0
3 years ago
Two bodies are falling with negligible air resistance, side by side, above a horizontal plane. If one of the bodies is given an
tankabanditka [31]

Answer:4-strikes the plane at same time as the other body

Explanation:

Given

If both bodies is falling on a horizontal plane and second body is given an acceleration in horizontal direction then it does not change the time to reach the Horizontal Plate as there is no change in vertical direction.

Horizontal acceleration will give only horizontal range and horizontal velocity.

8 0
3 years ago
Read 2 more answers
During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa
maw [93]

Answer:

Explanation:

Given

Initial speed u=60\ mi/hr\approx 88\ ft/s

distance traveled before coming to rest d_1=120\ ft

using equation of motion

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

0-(88)^2=2\times a\times 120---1

for u_2=80\ mi/hr\approx 117.33\ ft/s

using same relation we get

0-(117.33)^2=2\times a\times (d_2)----2

divide 1 and 2 we get

(\frac{88}{117.33})^2=\frac{120}{d_2}

d_2=213.32\ ft

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0
3 years ago
To prevent accidental electric shock, most electrical equipment is A. painted. B. isolated. C. coated. D. grounded.
Nesterboy [21]
D. It is Grounded below ground 
4 0
3 years ago
Read 2 more answers
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