2 years ago

Keeping in mind the kinetic energy of a moving vehicle, how can a driver best prepare to enter sharp curves in the roadway

Physics

2 answers:

777dan777 [17]2 years ago

8 0

Gradually slow down to a safe speed well before entering the curve.

elena-14-01-66 [18.8K]2 years ago

5 0

All u need to do is slow down

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Which of the following factors affects the pressure of an enclosed gas?

melamori03 [73]

It would be d all of the above

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3 years ago

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A soccer player kicks a soccer ball initially at rest setting it in motion at a velocity of 30 m/s. If the ball has a mass of 0.

Alchen [17]

Answer:

F= 600 N

Explanation:

Given that

Initial velocity ,u= 0 m/s

Final velocity ,v= 30 m/s

mass ,m = 0.5 kg

time ,t= 0.025 s

The change in the linear momentum is given as

ΔP= m (v - u)

ΔP= 0.5 ( 30 - 0 ) kg.m/s

ΔP= 15 kg.m/s

We know that from second law of Newtons

$F=\dfrac{dP}{dt}$

$F=\dfrac{\Delta P}{t}$

Now by putting the values

$F=\dfrac{15}{0.025}\ N$

F= 600 N

5 0

2 years ago

A 46.5-kg ball has a momentum of 57.2 kg m/s. What is the ball's speed?

Serhud [2]

Answer:

1.23 m/s

Explanation:

p=mv

57.2 = 46.5v

v= 57.2/46.5

v= 1.23

If you want to verify your answer, just insert the value of v in the equation.

5 0

2 years ago

Consider a highway composed of concrete. If the road is 4.5 km long, 30 m wide, and 0.70 m thick, what is its mass?

Aloiza [94]

Answer:

Explanation:

We need to assume that the density of the concrete is about 2350 Kg/m^3. And using the dimensions of the highway we can calculate the volume of the highway.

$vh=4500 * 30 * 0,70\\vh=94500 m^3\\den=m/v\\m=den*v\\m=2350*94500\\\\m=222075 ton$

5 0

2 years ago

The vertical force f acts downward at A on the two membered frames. Determine the magnitude of the two components of F directed

galben [10]

Answer:

The magnitude will be "353.5 N". A further solution is given below.

Explanation:

The given values is:

F = 500 N

According to the question,

In ΔABC,

⇒ $\angle BCA = (90-30)$

⇒ $=60^{\circ}$

then,

⇒ $\angle BAC=(180-45-60)$

⇒ $=75^{\circ}$

Now,

The corresponding angle will be:

⇒ $\angle FAC=60^{\circ}$

⇒ $\angle FAB=70+60$

⇒ $=135^{\circ}$

Aspect of F across the AC arm will be:

= $F\times cos(60)$

On putting the values of F, we get

= $500\times (.5)$

= $200 \ Newton$

Component F along the AC (in magnitude) will be:

= $F\times cos(135)$

= $500\times (-.707)$

= $-353.5 \ N \$

4 0

2 years ago