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MA_775_DIABLO [31]
4 years ago
10

On sunday a strong gale blows near beth’s house. the best estimate for the wind speed is ____ miles/hour.

Physics
1 answer:
Rufina [12.5K]4 years ago
8 0

Answer:

B. 47-54 miles/hour

Explanation:

Gale is a strong wind which is depicted by red warning flag. According to U.S. National Weather Service gale is a sustained surface wind. It is also used to refer winds from tropical coastal areas.

On the basis of force of wind gale is divided in four groups:

  1. Near gale   - 32-38 mph
  2. Gale            - 39-46 mph
  3. Strong gale - 47-54 mph
  4. Storm           - 55-63 mph
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The focal length of the given spherical mirror is 20cm.

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John accidentally drops his keys off the balcony at his apartment. John's friend Tony just happens to walk by at that moment and
EleoNora [17]

Explanation:

In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:

H = V²Sin²θ/2g

where,

V = Launch Speed = 18 m/s

θ = Launch Angle = 40°

g = 9.8 m/s²

Therefore,

H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)

H = 6.83 m

Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).

Therefore, the keys will make it back to John.

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When have you experienced an increase in kinetic<br> energy within a system?
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If a man starts running on a boat with an acceleration a with respect to the boat, there is no external force that acts on the Boat+Man system

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A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.
ValentinkaMS [17]

A) 2.03 m/s^2

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

mg sin \theta - \mu_k R = ma (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

\theta=22.5^{\circ}

\mu_k = 0.19 is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

R-mg cos \theta =0 (2)

From (2) we find

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2

B) 5.94 m/s

We can solve this part by using the suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

a=2.03 m/s^2

s = 8.70 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s

6 0
3 years ago
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