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3 years ago

Active transport is a process that uses energy to move materials through a

Physics

1 answer:

Nat2105 [25]3 years ago

5 0

Not good with this but I’m gonna guess cell membrane?

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an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

storchak [24]

<span>On what:

f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$
$\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}$
$\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}$
$\frac{1}{f} = \frac{10}{33.18}$
Product of extremes equals product of means:
$10*f = 1*33.18$
$10f = 33.18$
$f = \frac{33.18}{10}$
$\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark$

6 0

3 years ago

The conversion of thermal energy into mechanical energy requires

emmainna [20.7K]

Temperature difference is required, so i’m guessing - a. thermometer - would be required to check that temperature.

8 0

3 years ago

Rank the objects below in order from smallest to largest, where 1 is the smallest and 5 is the largest. local group of galaxies

sasho [114]

1. earth
2. solar system
3. milky way galaxy
4. local group of galaxies
5. universe

4 0

3 years ago

A violin string E vibrates faster then another violin string G when bowed. Which string produced a sound with higher frequency?

Lisa [10]

Violin string E

!!!!!!!!!!!!!!

5 0

2 years ago

Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it

stich3 [128]

Answer:

a. 6.69m/s

b. y=4.48m

c. t=1.43secs

Explanation:

Data given, acceleration,a=35m/s^2

distance covered,d=64cm=0.64m,

a. to determine the speed, we use the equation of motion

initial velocity,u=0m/s

if we substitute values we arrive at

$v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\$

b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2

and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.

Hence we can write the equation above again

$v^{2}=u^{2}-2a(y-2.2)\\$

if we substitute values we have

$v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m$

c. the time it takes to arrive at 1.83m is obtain by using the equation below

$1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37$

if we insert the values, we solve for t , hence t=1.43secs

6 0

3 years ago