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just olya [345]
3 years ago
8

A concave mirror forms a real image at 15.0 cm from the mirror surface along the principal axis. If the corresponding object is

at a distance of 30.0 cm, what is the mirror's focal length? Please show your work
Physics
1 answer:
kipiarov [429]3 years ago
6 0

Answer:

Focal length, f = 10 cm

Explanation:

We have,

A concave mirror forms a real image at 15.0 cm from the mirror surface along the principal axis, it means that the image distance is 15 cm

The object is placed at a distance of 30 cm, u = -30 cm

It is required to find the focal length of the mirror. Let it is f. Using mirror's formula to find it:

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{15}-\dfrac{1}{(-30)}=\dfrac{1}{f}\\\\f=10\ cm

So, the focal length of the mirror is 10 cm.

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Why can't 41 be represented in a binary table?
Airida [17]

Answer:

I think it is but I don't know for sure

Explanation:

41 101001

41 is 101001 on the binary table i think

8 0
3 years ago
A frictionless pulley used to lift 8000N of concrete. What is the minimum effort required to raise the block
rusak2 [61]

Answer: 8000N

Explanation: since it is frictionless that means it has 100% efficiency therefore the mechanical advantage is 1 meaning the load equals to the effort

3 0
3 years ago
I need help plz and thank you this is due
xeze [42]

Answer:

Grow up man, this is completely based on your curriculum, we would need your book to answer, and this has to be done by you.

3 0
3 years ago
Read 2 more answers
In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the ancho
melisa1 [442]

Answer:

largest lead = 3 m

Explanation:

Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.

So, first we need to calculate the velocities of both the anchorman  

given data:

Distance = d = 100 m

Time arrival for A = 9.8 s

Time arrival for B = 10.1 s

Velocity of anchorman A = D / Time arrival for A

=100/ 9.8 = 10.2 m/s

Velocity of anchorman B = D / Time arrival for B

=100/10.1 = 9.9 m/s

As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use

d = vt

= 9.9 x 9.8 = 97 m  

So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.

largest lead = 100 - 97 = 3 m

So if his lead no more than 3 m anchorman A win the race.

5 0
3 years ago
MATHPHYS HELP
rusak2 [61]

Answer:

16.4287

Explanation:

The force and displacement are related by Hooke's law:

F = kΔx

The period of oscillation of a spring/mass system is:

T = 2π√(m/k)

First, find the value of k:

F = kΔx

78 N = k (98 m)

k = 0.796 N/m

Next, find the mass of the unknown weight.

F = kΔx

m (9.8 m/s²) = (0.796 N/m) (67 m)

m = 5.44 kg

Finally, find the period.

T = 2π√(m/k)

T = 2π√(5.44 kg / 0.796 N/m)

T = 16.4287 s

3 0
3 years ago
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