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3 years ago

A box weighing 210 N is pushed up an inclined plane that is 2.0 m long. A force of 140 N is required. If?

Physics

2 answers:

Serhud [2]3 years ago

8 0

Answer:

Explanation:

N76 [4]3 years ago

3 0

D
Step-by-step explanation:

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amm1812

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3 years ago

An LED operation at 850 nm center wavelength has a spectral width of 45 nm. What is the pulse spreading in ns/km

Papessa [141]

Answer:

$\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}$

Explanation:

From the given information, the LED is operating with a given wavelength of 850 nm or 0.85 μm.

Hence, the material dispersion is $\dfrac {d \tau _{mat}}{d \lambda } \simeq (80 \ ps / (nm.km) \ )$

Now, using the pulse spread formula:

$\dfrac{\sigma_{mat}}{L} = \dfrac{d \tau _{mat} }{d \lambda} \sigma \lambda$

$\dfrac{\sigma_{mat}}{L} = (80 \ ps/ ( m.km) \ ) \times (45 \ nm)$

Thus, the pulse spreading as a result of material dispersion is: $\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}$

3 0

3 years ago

Enter the following expression in the answer box below: 2gλ3m−−−−√, where λ is the lowercase Greek letter lambda.

atroni [7]

Answer:

$\sqrt{\dfrac{2g\lambda^3}{m}}$

Explanation:

We can write the expression here, but the point of the problem seems to be to see if you can manipulate the controls on the answer box to reproduce that expression.

$\boxed{\sqrt{\dfrac{2g\lambda^3}{m}}}$

7 0

3 years ago

3. A cat walks 0.220km North, then 0. 120 km South in a time of 400 seconds. whats the displacement and average velocity?

Andru [333]

Answer:

The rate at which velocity changes with respect to a change in time is called. acceleration.

Explanation:

4 0

3 years ago

As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o

astraxan [27]

Answer:

3.6 x 10⁶ Pa

Explanation:

A = Area of the heel = 1.50 cm² = 1.50 x 10⁻⁴ m²

m = mass of the woman = 55.0 kg

g = acceleration due to gravity = 9.8 m/s²

Force of gravity on the heel is given as

F = mg

Inserting the values

F = (55) (9.8)

F = 539 N

Pressure exerted on the floor is given as

$P = \frac{F}{A}$

$P = \frac{539}{1.5\times 10^{-4}}$

P = 3.6 x 10⁶ Pa

6 0

3 years ago