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murzikaleks [220]
2 years ago
12

You are traveling 70 mph (31.3 m/s) and slam on the brakes to avoid hitting another car. How far do you travel if it takes you 8

.5 seconds to stop?
Physics
1 answer:
Arlecino [84]2 years ago
6 0

You traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

<u>Given the following data:</u>

  • Initial velocity, U = 31.3 m/s  
  • Time, t = 8.5 seconds.

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find the distance traveled, we would use the second equation of motion:

Mathematically, the second equation of motion is given by the formula;

S = ut + \frac{1}{2} at^2

Where:

  • S is the distance travelled.
  • u is the initial velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

Substituting the parameters into the formula, we have;

S = 31.3(8.5) + \frac{x}{y} (9.8)(8.5^2)\\\\S = 266.05 + 4.9(72.25)\\\\S = 266.05 + 354.025

<em>Distance, S</em><em> = </em><em>620.075 meters.</em>

Therefore, you traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

Read more: brainly.com/question/8898885

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The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm
Nesterboy [21]

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answer:

225 N

Explanation:

From Pascal's principle,

F/A = f/a ...................... Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A ..................... Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

8 0
3 years ago
A heavy truck and a small truck roll down a hill. Neglecting friction, at the bottom of the hill the heavy truck has greater
postnew [5]

Answer:

kenetic energy

Explanation:

or potential energy

3 0
3 years ago
A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball i
scoray [572]

Answer:

The distance traveled by the woman is 34.1m

Explanation:

Given

The initial height of the cliff

yo = 45m final, positition y = 0m bottom of the cliff

y = yo + ut -1/2gt²

u = 20.0m/s initial speed

g = 9.80m/s²

0 = 45.0 + 20×t –1/2×9.8×t²

0 = 45 +20t –4.9t²

Solving quadratically or by using a calculator,

t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s

So this is the total time it takes for the ball to reach the ground from the height it was thrown.

The distance traveled by the woman is

s = vt

Given the speed of the woman v = 6.00m/s

Therefore

s = 6.00×5.69 = 34.14m

Approximately 34.1m to 3 significant figures.

3 0
3 years ago
A physicist hangs a 150-g object on a spring whose spring constant is a value of 13.22 Newtons/meter and has a spring force of 2
Nataly_w [17]

Answer:

so the answer is this because the answer is that

Explanation:

and the reason why the answer is this and that is because the answer is that

4 0
2 years ago
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
slega [8]
We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
dI =r^2*dm
where dm =M(b*dr)/(ab)
Now we find the moment of inertia by integrating from -a/2 to a/2
The moment of inertia is
I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3) (from (-a/2) toI=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12 (a/2))



4 0
3 years ago
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