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just olya [345]
3 years ago
11

Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 34.5 m/s sinking curve ball head on, sending it of

f his bat in the exact opposite direction at 23.9 m/s. The actually contact between ball and bat lasted for 0.00075 seconds. Determine the magnitude of the average acceleration of the ball during the contact with the bat in m/s/s. Do not include units or /- sign. Round your answer to a whole number.
Physics
1 answer:
Aneli [31]3 years ago
3 0

Answer:

–77867 m/s/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration = (final velocity – Initial velocity) /time

a = (v – u) / t

With the above formula, we can obtain acceleration of the ball as follow:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

a = (v – u) / t

a = (–23.9 – 34.5) / 0.00075

a = –58.4 / 0.00075

a = –77867 m/s/s

Thus, the acceleration of the ball is –77867 m/s/s.

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Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

F = \dfrac{kq_1q_2}{r^2}

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8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}

Force due to Q₃ :

F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}

8.3. The net force on the particle at Q₂ is the vector

\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}

Its magnitude is

\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}

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The following are the Earth–Sun distance at the equinoxes and solstices: March equinox 149.0 million km June solstice 152.0 mill
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Answer:

During <u>winter (late December/early January)</u> the Earth is closest to the Sun and during <u>summer (late June/early July)</u> the Earth is farthest from the Sun.

Explanation:

In the northern hemisphere, the earth usually comes closer to the sun during the time of winter season, mostly in late December or early January.

On the other hand, the earth is farthest from the sun during the time of summer season, mostly in late June or early July.

When the earth is closer to the sun, during the winter, it is comparatively cold. It is due to the absorption of a lesser amount of incoming solar radiation. The tilt of the earth is also responsible for this low temperature.

But, when the earth is farthest from the sun, during the summer, it is comparatively hot. It is due to the absorption of a large amount of incoming solar radiation.

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The driver of a car slams on the brakes, causing the car to slow down at a rate of
Alona [7]

Answer:

The time taken for the car to stop is 5.43 s.

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Explanation:

The following data were obtained from the question:

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Distance travalled (s) = 295 ft

Final velocity (v) = 0 ft/s

Time taken (t) =?

Initial velocity (u) =?

Next, we shall determine the initial velocity of the car as shown below:

v² = u² + 2as

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0 = u² + (–11800)

0 = u² – 11800

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0 + 11800 = u²

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u = √11800

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Therefore, the initial velocity of the car is 108.6 ft/s.

Finally, we shall determine the time taken for the car to stop as shown below:

Acceleration (a) = –20 ft/s² (since the car is coming to rest)

Final velocity (v) = 0 ft/s

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Time taken (t) =?

v = u + at

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0 = 108.6 + (–20t)

0 = 108.6 – 20t

Collect like terms

0 – 108.6 = – 20t

– 108.6 = – 20t

Divide both side by –20

t = – 108.6 / –20

t = 5.43 s

Therefore, the time taken for the car to stop is 5.43 s.

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