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cupoosta [38]
3 years ago
15

Which describes an image that can be produced by a concave lens?

Physics
1 answer:
mixer [17]3 years ago
5 0
A concave is when the bump is pressed downwards; So if the image has a downward hole, then it's concave. If it's pressed upwards like a hill or mountain, then it's convex.
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A 64 kg swimmer jumps, with a velocity of 4.2 m/s, off the front of a 25 kg kayak when the kayak is moving forward at a velocity
Crank

Answer:

3.88m/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and 2 are the initial velocities

v is the final velocity

Given

m1 = 64kg

u1 = 4.2m/s

m2 = 25kg

u2 = 3.2m/s

Required

Final velocity v

Substitute the given values into the formula

64(4.2)+25(3.2) = (65+25)v

268.8+80 = 90v

348.8 = 90v

v = 348.8/90

v = 3.88m/s

Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s

8 0
3 years ago
Which formation is one feature of karst topography?
steposvetlana [31]
Karst is a topography formed from the dissolution of soluble rocks such as limestone, dolomite, and gypsum. It is characterized by underground drainage systems with sinkholes and caves. It has also been documented for more weathering- resistant rocks, such as quartzite, given the right conditions.
8 0
3 years ago
Read 2 more answers
Use the dimensional analysis and check the correctness of given equation:-<br> PV= nRT
Harman [31]

PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

\\ \tt\bull\leadsto PV

\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]

\\ \tt\bull\leadsto [ML^5T^{-2}]

RHS

\\ \tt\bull\leadsto nRT

\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]

\\ \tt\bull\leadsto [ML^5T^{-2}]

LHS=RHS

hence verified

6 0
3 years ago
Read this excerpt from Through the Looking-Glass by Lewis Carroll.
Vanyuwa [196]

Answer:

Why does Alice forget the name of the woods and her own name?

6 0
3 years ago
An 87.6 g lead ball is dropped from rest from a height of 7.00 m. The collision between the ball and the ground is totally inela
bija089 [108]

Taking specific heat of lead as 0.128 J/gK = c

We have energy of ball at 7.00 meter height = mgh = 87.6*10^{-3}*9.81*7

When leads gets heated by a temperature ΔT energy needed = mcΔT

                                                                      = 87.6*10^{-3}*0.128*10^3ΔT

Comparing both the equations

                      87.6*10^{-3}*9.81*7 = 87.6*10^{-3}*0.128*10^3ΔT

                        ΔT = 0.536 K

                        Change in temperature same in degree and kelvin scale

                                      So ΔT = 0.536 ^0C

7 0
3 years ago
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