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3 years ago

Not sure what to put this under

Physics

1 answer:

cestrela7 [59]3 years ago

5 0

family 16 cause i said so XD

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An object moves along the x axis. The graph shows it’s position x as a function of time t. Find the average velocity of the obje

nadezda [96]

Question is from B to C

Answer: (b) 1.5m/s

x1=3m, x2=9m
t1=1s, t2=5s
Displacement, ∆x=(9-3)m=6m
Time elapsed, ∆t=(5-1)s=4s

So average velocity v =∆x/∆t=6/4=1.5m/s

6 0

3 years ago

A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm

iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

$m = m_{o}* m_{e}$

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

$m_{o} = \frac{v_{o}}{-u _{0}}$

and me as

$m_{e} = 1+\frac{D}{f_{e}}$

d is distant of distinct vision = 25.0 cm for normal eye

fe = focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}$

$\frac{1}{v_{o}} = 2.1 cm$

$m_{o} = \frac{2.1}{0.150} = 14$

$m_{e} = 1+\frac{25}{1}$

$m_{e} =26$

$m = m_{o}* m_{e}$

m = 14*26 = 364

4 0

3 years ago

An electric car is being designed to have an average power output of 4,600 W for 2 h before needing to be recharged. (Assume the

MAVERICK [17]

Answer:

a)3312 x 10⁴ J

b)I = 57.5 A

c)9200 W

Explanation:

Given that

P =4600 W

Time t= 2 h = 2 x 3600 s= 7200 s

We know that

1 W = 1 J/s

Energy stored in the battery = P .t

=4600 x 7200 J

=3312 x 10⁴ J

We know that power P given as

P = V .I

V=Voltage ,I =Current

4600 = 80 x I

I = 57.5 A

The energy supplied = 4600 x 2 = 9200 W

7 0

3 years ago

Which scenario would result in a decrease in population size, or a negative population growth?

Anestetic [448]

I’m pretty sure the answer is b

5 0

3 years ago

Read 2 more answers

An isloated point charge produce an electric field with magnitude E at a point 2 m away. At a point 1 m from the charge magnitud

Nina [5.8K]

Answer:

the correct answer is C, E’= 4E

Explanation:

In this exercise you are asked to calculate the electric field at a given point

E = $k \frac{q}{r^2}$

indicates that the field is E for r = 2m

E = $\frac{ k q}{4}$ (1)

the field is requested for a distance r = 1 m

E ’= k \frac{q}{r'^2}

E ’= k q / 1

from equation 1

4E = k q

we substitute

E’= 4E

so the correct answer is C

8 0

3 years ago