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Lyrx [107]
3 years ago
7

While driving on a highway, your HMMWV starts to pull to one side. You stop and check your tires – one tire is going flat. There

is no spare, but the affected tire has a runflat device installed, and you are only 5 miles from the motor pool. What is your maximum allowable speed when driving with one flat tire?
Physics
1 answer:
MrRissso [65]3 years ago
3 0

Answer:

30 miles people hour (mph)

Explanation:

The maximum speed for one flat tyre is 30mph, for both tyres it is 20 mph and for two tyres on the same side it is 30mph. When a tyre is punctured driving at high speed should be avoided and not too long distance should be covered to avoid further accident and for the safety of other vehicles. Speed limit and distance for flat tires should be strictly followed.

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Developed countries typically have _______.
Anna35 [415]

Answer:

Good government

Explanation:

Many people will think its natural resources but its not. Africa is a proof. They have resources but leadership is what makes most of their countries 3rd world. #BADGOVERNANCE

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2 years ago
Which image illustrates refraction
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Answer:

it loaded and it is C. buddy sorry about that :)

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3 years ago
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A 2.0-kg mass is projected from the edge of the top of a 20-m tall building with a velocity of 24 m/s at some unknown angle abov
Digiron [165]

Ox:vₓ=v₀

x=v₀t

Oy:y=h-gt²/2

|vy|=gt

tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g

y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°

cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s

Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ

3 0
2 years ago
A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit
Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

y = \frac{1}{2} at^2+v_0t+y_0

Where,

a = acceleration

t = time

v_0 = Initial velocity

y_0 = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

v = \frac{x}{t}

x = Displacement

t = time

With the data we have we can find the time as well

v = \frac{x}{t}

t = \frac{x}{v}

t = \frac{18.6}{34}

t = 0.547s

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

y = \frac{1}{2} at^2+v_0t+y_0

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y = \frac{1}{2} gt^2

y = \frac{1}{2} 9.8*0.547^2

y = 1.46m

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0
3 years ago
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