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klio [65]
3 years ago
14

Difference between inverted and upright microscope

Physics
1 answer:
ser-zykov [4K]3 years ago
6 0
An Inverted Microscope gives you more freedom than an upright microscope. Inverted microscopes are very useful to examine the surface of heavy and large sized Items for industrial purposes. Whereas <span>upright microscopes have very limited distance between the table and the objective.</span>
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A closed pipe creates a
lina2011 [118]

Answer:

375 hz

Explanation:

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5 0
3 years ago
The wavelengths of radio waves are much _______________ than the wavelength of microwaves. therefore, radio waves carry much ___
nydimaria [60]
The wavelengths of radio waves are much "Longer" than the wavelength of microwaves therefore, radio waves carry much "Lower" <span>energy than a microwave.

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7 0
3 years ago
Read 2 more answers
George and Harriot walk with an average velocity of .95 m/s eastward. If it takes them 30
artcher [175]

Answer:

1.71 km

Explanation:

Convert 30 minutes to seconds:

30 min × (60 s / min) = 1800 s

Find the displacement:

0.95 m/s × 1800 s = 1710 m

Convert to kilometers:

1710 m × (1 km / 1000 m) = 1.71 km

5 0
3 years ago
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A ball is thrown straight up with enough speed so that it is in the air for several seconds. Assume the positive direction is up
Andrew [12]

Answer:

a) v= 0 m/s b) v= 6.86 m/s

Explanation:

a) When the ball reaches to its highest point, under the influence of gravity, before starting to fall down, it momentarily comes to an stop (this is needed prior to change direction in any movement), so, applying the definition of acceleration, and replacing the acceleration a by g, we have:

vf = v₀ - g*t (1)

The minus sign means that the acceleration due to gravity is always downward, so if we assume that the positive direction is upwards it must be negative.

At the highest point, vf= 0.

b) Prior to solve this point, we need to know which is the time when the ball reaches to its highest point.

As we know vf=0, we can solve (1) for t, as follows:

th = v₀/g

Now, for a time that is 0.7 s before this time, applying the acceleration definition and solving for v again, we have:

v = v₀ -(g *(th-0.7 s)), but th= v₀/g, so we get:

v= v₀ -g((v₀/g)-0.7 s) = v₀ - v₀ + g*0.7 s

⇒ v=g*0.7 s = 9.8 m/s²*0.7 s

⇒ v = 6.86 m/s

6 0
3 years ago
A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/
s344n2d4d5 [400]

Answer:

At t = (70 / 3) \; {\rm s} (approximately 23.3 \; {\rm s}.)

Explanation:

Note that the acceleration of the car between t = 0\; {\rm s} and t = 20\; {\rm s} (\Delta t = 20\; {\rm s}) is constant. Initial velocity of the car was v_{0} = 0\; {\rm m\cdot s^{-1}}, whereas v_{1} = 35\; {\rm m\cdot s^{-1}} at t = 20\; {\rm s}\!. Hence, at t = 20\; {\rm s}\!\!, this car would have travelled a distance of:

\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

In other words, at t = 20\; {\rm s}, the truck was 400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m} ahead of the car. The velocity of the car is greater than that of the truck by 35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}. It would take another (50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s} before the car catches up with the truck.

Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

3 0
2 years ago
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