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2 years ago

Pilots often take advantage of the ____,which are highly-speed winds between 7km and 16 km above earths surface.

1 answer:

6 0

I believe it is called or referred to as the "Jet Stream". During World War II, allied pilots encountered high speed winds in the upper air. They named those winds after the fastest planes they came up against: fighters equipped with jet engines! Jet stream winds in winter time can reach up to 300 MPH as well!

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A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is

Aleksandr [31]

Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

Σ $F_{y}$ =0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

$N-W=0$

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

$f_{k}=N$ μ

where μ is the kinetic friction coefficient.

So we get that the kinetic friction is:

$f_{k}=2450N*0.12$

$f_{k}=294$

With this information we can go ahead and find the sum of horizontal forces:

Σ $F_{x}$ =ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N- $f_{k}$ =ma

750N-294N=(250kg)a

when solving for a we get:

$a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}$

so the crate's acceleration is 1.82m/s².

5 0

2 years ago

PLSS HELP

trasher [3.6K]

Answer:

Wavelength and Frequency

8 0

2 years ago

Read 2 more answers

A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can r

mr Goodwill [35]

Answer:

Explanation:

Let T be the tension .

Applying newton's second law on the downward movement of the bucket

mg - T = ma

On the drum , a torque of TR will be acting which will create an angular acceleration of α in it . If I be the moment of inertia of the drum

TR = Iα

TR = Ia/ R

T = Ia/ R²

Replacing this value of T in the other equation

mg - T = ma

mg - Ia/ R² = ma

mg = Ia/ R² +ma

a ( I/ R² +m)= mg

a = mg / ( I/ R² +m)

mg - T = ma

mg - ma = T

mg - m x mg / ( I/ R² +m) = T

mg - m²g / ( I/ R² +m ) = T

mg - mg / ( 1 + I / m R² ) = T

b ) T = Ia/ R²

I = TR² / a

c ) Moment of inertia of hollow cylinder

I = 1/2 M ( R² - R² / 4 )

= 3/4 x 1/2 MR²

= 3/8 MR²

I / R² = 3/8 M

a = mg / ( I/ R² +m)

a = mg / ( 3/8 M + m )

T = Ia/ R²

= 3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²

= $\frac{3mMg}{(3M +8m)}$

7 0

3 years ago

A rifle is fired and recoils when the bullet leaves the gun. This is an example of Newton's third law. The acceleration of the b

sweet [91]

<span>The acceleration of the bullet is called ACTION.

</span>Formally stated, Newton's third law<span> is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

5 0

2 years ago

Read 2 more answers

A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A (N/m²)

P1=P2

$\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }$

$F_{2} = \frac{F_{1}*A_{2} }{A_{1}}$ Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight (W₂) can the large piston support

We replace data in the equation (1)

$F_{2} = \frac{(500)*(40) }{2}$

F₂ = 10000 N

W₂= F₂= 10000 N

6 0

3 years ago