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ycow [4]
4 years ago
11

You throw a baseball directly upward at time t = 0 at an initial speed of 12.3 m/s. What is the maximum height the ball reaches

above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9.80 m/s2.
Physics
1 answer:
Semmy [17]4 years ago
5 0

Explanation:

At the maximum height, the ball's velocity is 0.

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (12.3 m/s)² + 2(-9.80 m/s²)(x - 0 m)

x = 7.72 m

The ball reaches a maximum height of 7.72 m.

The times where the ball passes through half that height is:

x = x₀ + v₀ t + ½ at²

(7.72 m / 2) = (0 m) + (12.3 m/s) t + ½ (-9.8 m/s²) t²

3.86 = 12.3 t - 4.9 t²

4.9 t² - 12.3 t + 3.86 = 0

Using quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 12.3 ± √(12.3² - 4(4.9)(3.86)) ] / 9.8

t = 0.368, 2.14

The ball reaches half the maximum height after 0.368 seconds and after 2.14 seconds.

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If the magnitude of the electric field in air exceeds roughly 3 ✕ 106 n/c, the air breaks down and a spark forms. for a two-di
Vlad1618 [11]

Answer: 39.8 μC

Explanation:

The magnitude of the electric field generated by a capacitor is given by:

E = \frac{V}{d}

d is the distance between the plates.

For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.

C =\frac{\epsilon_o A }{d}

where A is the area of the plate and ε₀ is the absolute permittivity.

substituting, we get

E = \frac{Q}{\epsilon_o A}

It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.

radius of the plates of the capacitor, r = 69 cm = 0.69 m

Area of the plates, A = πr² = 1.5 m²

Thus, the maximum charge that can be placed on disks without a spark is:

Q = E×ε₀×A

⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.

8 0
3 years ago
Solving:
STatiana [176]

Answer:

1. 31,536,000 seconds

2. Car B traveled a longer distance

3. Volume of box = 0.235887 cubic meters

Explanation:

Q1. Age in seconds

1 year = 365 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds

Therefore 1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds

In scientific notation this would be 3.1536\times  10^7 \text{ seconds}

Q2. Comparing km and miles

Given:
1 km = 1000m and 1 m = 3.3ft,   I km = 1000 x 3.3 = 3300 ft.

Convert Car A distance of 25.7km to feet :
25.7 km. = 25. 7 x 3300 ft. = 84,810 ft.

For Car B  that traveled 20 miles,
20 miles = 20 x 5280 = 105,600 ft.

Since 105,600 > 84,810, car B traveled a longer distance

Q3. Volume of wooden box

The wooden box is in the shape of a rectangular prism
It volume is L x W x H
Volume = 1.525 x 0.30 x 0.5156 = 0.235887 cubic meters

5 0
1 year ago
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Xelga [282]

Answer:

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Explanation:

y - dirdf leter of di alfabet

3 0
3 years ago
How can a large construction project impact water quality, supply or flow?
Korolek [52]

Answer:

Construction also impacts the community’s water quality. During the construction phase, pollutants can end up in the water supply. Once construction is over, the water quality can still be a problem due to an increased demand for water, as well as increased pollution due to additional residents.

Explanation:

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3 0
3 years ago
2 kg steel ball is dropped straight down ont a hard horixontal florr and bounces straight up . the balls speed just before and j
Rainbow [258]

Answer:

40 \frac{kg*m}{s}

Explanation:

Impulse-momentum theorem states that impulse is equal to the change of momentum:

\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i} (1)

with pf the final momentum and pi the initial momentum. Knowing that momentum is mass (m) times velocity (1) is:

\overrightarrow{J}=m \overrightarrow{v}_{f}- m \overrightarrow{v}_{i}

It's important to note that we're dealing with vector quantities so direction matters. If we choose towards the floor positive direction then the initial velocity is positive and the final velocity is negative, so:

\overrightarrow{J}=-(2kg)(10\frac{m}{s}) - (2.0kg)(10\frac{m}{s})

\overrightarrow{J}=-40 \frac{kg*m}{s}

So, the impulse delivered to the floor is 40 \frac{kg*m}{s}

5 0
3 years ago
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