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jek_recluse [69]
2 years ago
5

The driver then tests the brakes on the car and safely comes to a complete stop with constant acceleration from 26.8 meters per

second in only 32.3 meters.
A. What is the magnitude and direction of the cars acceleration?

B. How much time did the car travel under this constant acceleration before stopping?
Physics
1 answer:
Temka [501]2 years ago
5 0

Answer:

62.78

Explanation:

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Q. At what point in a waterfall do the drops of water contain the most kinetic energy ?
antoniya [11.8K]

Answer:

2.When they reach the bottom of the fall

Explanation:

The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with  decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).

Thus, the correct option is "2" When they reach the bottom of the fall

7 0
3 years ago
A rod 7.0 m long is pivoted at a point 2.0 m from the left end. A downward force of 50 N acts at the left end, and a downward fo
kicyunya [14]

If the rod is in rotational equilibrium, then the net torques acting on it is zero:

∑ τ = 0

Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:

• at the left end,

τ = + (50 N) (2.0 m) = 100 N•m

• at the right end,

τ = - (200 N) (5.0 m) = - 1000 N•m

• at a point a distance d to the right of the pivot point,

τ = + (300 N) d

Then

∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0

⇒   (300 N) d = 1100 N•m

⇒   d ≈ 3.7 m

6 0
2 years ago
Who water rocket starts from rest and roses straight up with an acceleration of 5 m/s until it runs out of water 2.5 seconds lat
Kitty [74]

Answer:

23. 4375 m

Explanation:

There are two parts of the rocket's motion

1 ) accelerating  (assume it goes upto  h1 height )

using motion equations upwards

s = ut+\frac{1}{2}*a*t^{2} \\h_1=0+\frac{1}{2}*5*2.5^{2} \\=15.625 m

Lets find the velocity after 2.5 seconds (V1)

V = U +at

V1 = 0 +5*2.5 = 12.5 m/s  

2) motion under gravity (assume it goes upto  h2 height )

now there no acceleration from the rocket. it is now subjected to the gravity

using motion equations upwards (assuming g= 10m/s² downwards)

V²= U² +2as

0 = 12.5²+2*(-10)*h2

h2 = 7.8125 m

maximum height = h1 + h2

                            = 15.625 + 7.8125

                            = 23. 4375 m

3 0
3 years ago
If the element gallium has an atomic number of 31 and an atomic mass of 70, how many neutrons does it have?
Mumz [18]

Answer:

The answer is 39.

Explanation:

The atomic number refers to the number of protons and the atomic mass is the sum of the protons and neutrons. So, you would just do 70 - 31 and that gets you 39.

6 0
2 years ago
A spring stretches 1.68cm vertically when a 2.50kg object is suspended from it.Find the distance (in cm) the spring stretches if
Llana [10]

Answer:

2.96 cm

Explanation:

By Hook's law

Force(F) = Spring constant(k) × Extension(d)

 F = k × d

Force is the weight of the object, F = W = mg

So we get, mg = kd ⇒ m ∝ d

                                   2.5 ∝ 1.68  --------------(1)

                                   4.4 ∝ d'      --------------(2)

From (1) & (2),     4.4/2.5 = d'/1.68

                                   d' = 2.96 cm ⇒ the required extension.

7 0
3 years ago
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