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3 years ago

What is the complete cycle of ice wedging

Physics

1 answer:

timofeeve [1]3 years ago

3 0

Water works it way through a crack then it freezes

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Electro-magnets are created by...

mojhsa [17]

Wrapping in insulated wire around a medal with ferromagnetic properties and applying an electric current

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3 years ago

An opera singer in a convertible sings a note at 600 Hz while cruising down the highway at 90km/hr. What is the frequency heard

Inessa05 [86]

Answer:

$f=647.17Hz$

Explanation:

Given data

Frequency fs=600 Hz

Velocity Vs=90 km/hr =25 m/s

Speed of sound v=343 m/s

To find

Frequency f heard by a person standing beside the road in front of the car

Solution

From Doppler's effect we know that:

$f=f_{s}(\frac{v+vL}{v+vs} )\\ as\\vL=0$

Here Vs is negative as the source is moving towards listener

$f=fs(\frac{v+0}{v-vs} )\\f=(600Hz)(\frac{343m/s+0}{343m/s-25m/s} )\\f=647.17Hz$

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4 years ago

I WILL GIVE BRAINLIEST!!!!

solniwko [45]

Blue is pulling 100n harder

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3 years ago

Paco was driving his scooter west with an initial velocity of 4 m/s. He accelerates at 0.5 m/s2 for 30 seconds. What is his fina

Kipish [7]

Answer:

19m/s

Explanation:

U= 4m/s

a = 0.5m/s²

t= 30s

V= U+at

= 4+15

=19m/s

7 0

4 years ago

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti

Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0

3 years ago