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3 years ago

Why is a specimen smaller than 200 nm not visible with a light microscope?

Physics

1 answer:

kifflom [539]3 years ago

8 0

Because the specimen is very small with a light microscope

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An empty beaker is placed on a top-pan balance. Some water is now poured into the beaker.What is the weight of the water? A. 0.0

mario62 [17]

Answer:

A. 0.044 kg

Explanation:

We need to subtract the sum of (beaker+water - empty beaker) which is 0.106 kg - 0.062 kg = 0.044 kg. The answer will not be written in Newton because this unit is used for force only and in this question w have to find the weight.

Hope it is enough.

Please mark me as brainliest.

6 0

3 years ago

A skateboarder who travels 60 meters in 30 seconds has a speed of

natulia [17]

2m/s

Explanation:

Given parameters:

Distance traveled = 60m

time taken = 30seconds

Unknown:

Speed of skateboarder = ?

Solution:

Speed is the rate of change of distance with time taken. It is a scalar quantity that only revers magnitude;

Speed = $\frac{distance}{time}$

Speed = $\frac{60}{30}$ = 2m/s

learn more:

Speed brainly.com/question/1548911

#learnwithBrainly

4 0

3 years ago

A wheel of mass 4kg is pulled up a plane inclined at 30° to the horizontal by a force of 45N applied to the axle and parallel to

Len [333]

Answer:

v = 10 m/s

Explanation:

Let's assume the wheel does not slip as it accelerates.

Energy theory is more straightforward than kinematics in my opinion.

Work done on the wheel

W = Fd = 45(12) = 540 J

Some is converted to potential energy

PE = mgh = 4(9.8)12sin30 = 235.2 J

As there is no friction mentioned, the remainder is kinetic energy

KE = 540 - 235.2 = 304.8 J

KE = ½mv² + ½Iω²

ω = v/R

KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²

v = √(2KE / (m + I/R²))

v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6

v = 10.07968...

5 0

3 years ago

Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject

Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

$t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}$

where

$\mu_{sun}$ is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2. $t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}$

t = 12,105.96 sec

6 0

3 years ago

Help me pls, class starts soon

ratelena [41]

Answer:

Explanation:

1.)66.36

2.)11.4

3.)0.8104

4.)4158.315

i got some of them and try the rest good luck

4 0

3 years ago