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Katena32 [7]
3 years ago
14

a foul ball is hit into the stands at a baseball game. the ball rises to a height of 38 meters and is caught on its way down by

a fan standing in the bleaches 30 meters above the height at which it was hit
Physics
1 answer:
lisov135 [29]3 years ago
4 0

The velocity of the ball when it was caught is 12.52 m/s.

<em>"Your question is not complete it seems to be missing the following, information"</em>,

find the velocity of the ball when it was caught.

The given parameters;

maximum height above the ground reached by the ball, H = 38 m

height above the ground where the ball was caught, h = 30 m

The height traveled by the ball when it was caught is calculated as follows;

y = H - h

y = 38 - 30 = 8 m

The velocity of the ball when it was caught is calculated as;

v_f^2 = v_0 + 2gh\\\\v_f^2 = 0 + (2\times 9.8 \times 8)\\\\v_f^2 = 156.8\\\\v_f = \sqrt{156.8} \\\\v_f = 12.52 \ m/s

Thus, the velocity of the ball when it was caught is 12.52 m/s.

Learn more here: brainly.com/question/14582703

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A 240 V circuit has 20 A flowing through it. How much power is it using?
Delvig [45]

Explanation:

P = IV

V = 240

I = 20

P = ?

P = 240 × 20

P = 4800 watts

6 0
3 years ago
A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is
lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth =  5.98 × 10²⁴ kg

Radius of Earth = 6.38\times 10^6\ m

The gravitational force will balance the centripetal force

\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}

T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}

From Kepler's law we have relation

T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m

Distance from the center of the Earth would be

42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}

8 0
3 years ago
Provided following are four different ranges of stellar masses. Rank the stellar mass ranges based on how many stars in each ran
elena-s [515]

Highest to lowest number:

-less than 1 solar mass

-between 1 and 10 solar masses

-between 10 and 30 solar masses

-between 30 and 60 solar masses

<h3>What is Stellar masses ?</h3>

Stellar mass is a phrase that is used by astronomers to describe the mass of a star.

  • It is usually enumerated in terms of the Sun's mass as a proportion of a solar mass ( M ☉). Hence, the bright star Sirius has around 2.02 M ☉.

  • Stellar masses are not fixed, although they change for single stars only on long periods.

Learn more about Stellar masses here:

brainly.com/question/1128503

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2 years ago
Between which two points of a concave mirror should an object be placed to obtain a magnificati of -3​
kobusy [5.1K]

Answer:

When object is placed between the focus (F) and pole (P) of a concave mirror, magnified and erect image of the object is formed on the back of the mirror.

When object is placed between the centre of curvature and the principal focus of a concave mirror, magnified and inverted image is formed in front of the mirror.

Explanation:

8 0
3 years ago
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
3 years ago
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