PHYSICS COLLEGE

A baseball is hit with a speed of 47.24 m/s from a height of 0.42 meters. If the ball is in the air 5.73 seconds and lands 130 meters from the batters feet, (a) at what angle did the ball leave the bat? (b) with what velocity will the baseball hit the ground?

Answers

Answer 1

Answer:

Answer:

a)the ball will leave the bat at an angle of 61.3° .

b) the velocity at which it will hit the ground will be v = 27.1 m/s

Explanation:

Given,

v = 47.24 m

h = 0.42 m

t = 5.73 s

R = 130 m

a)We know that

R = v cosθ × t

cosθ = $(R)/(v t ) = (130)/(47.24* 5.73 ) =0.4803$

θ = 61.3°

the ball will leave the bat at an angle of 61.3° .

b)Vx = v cos(θ) = 47.24 x cos 61.3 = 22.7 m/s

v = u + at

Vy = 47.24 x sin 61.3 - 9.81 x 5.73

= -14.8 m/s

v = $√(v_x^2 + v_y^2))$

v = $√(22.7^2 + -14.8^2)$

v = 27.1 m/s

the velocity at which it will hit the ground will be v = 27.1 m/s

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Answers

Answer:

Explanation:

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A toy airplane is flying at a speed of 6 m/s with an acceleration of 0.3 m/s2How fast is it flying after 4 seconds?
A. 5.7 m/s
B. 2.6 m/s
C. 13.9 m/s
D. 7.2 m/s
SUBMIT

Answers

Answer:

Explanation:

0.3=v-6/4

make v the subject

v=7.2ms^2

The direct sunlight at Earth's surface is about 1050 W/m2 . Compute mass lost by Sun in a thousand years as a fraction of Earth Mass? The lost Mass of sun/in a million yrs = ……?…..% of Earth Mass.

Answers

Answer:

Explanation:

Energy falling on 1 m² surface of earth per second = 1050

Energy in one million years on 1 m²

= 1050 x 60 x 60 x 24 x 365 x 10⁶ = 3.311 x 10¹⁶ J

In order to calculate total energy coming out of the surface of the sun , we shall have to sum up this energy for the while spherical surface of imaginary sphere having radius equal to distance between sun and earth.

Area of this surface = 4π R² = 4 X 3.14 X (149.6 X 10⁹ )²

= 2.8 X 10²³ m²

So total energy coming out of the sun = 2.8 x 10²³ x 3.311 x 10¹⁶

= 9.271 x 10³⁹ J

From the formula

E = mc² { energy mass equivalence formula }

m = E / c² = $(9.271 *10^(39))/(9 * 10^(16))$

1.03 x 10²³ kg

mass of earth = 5.972 x 10²⁴

Answer in percentage of mass of earth

= $(1.03*10^23)/(5.972*10^(24))*100$

= 1.72 %

We showed that the length of the pendulum of period 2.000 seconds on the Earth’s surface was 0.99396 meters. What period would this same pendulum have on the surface of Mars? What length would the pendulum be in order to have a period of 2.000 seconds?

Answers

To solve this problem it is necessary to apply the concepts related to the Period based on gravity and length.

Mathematically this concept can be expressed as

$T= 2\pi \sqrt{(l)/(g)}$

Where,

l = Length

g = Gravitational acceleration

First we will find the period that with the characteristics presented can be given on Mars and then we can find the length of the pendulum at the desired time.

The period on Mars with the given length of 0.99396m and the gravity of the moon (approximately $1.62m / s ^ 2)$ will be

$T= 2\pi \sqrt{(l)/(g)}$

$T= 2\pi \sqrt{(0.99396)/(1.62)}$

$T = 4.921seg$

For the second question posed, it would be to find the length so that the period is 2 seconds, that is:

$T= 2\pi \sqrt{(l)/(g)}$

$2= 2\pi \sqrt{(l)/(1.62)}$

$l = 0.16414m$

Therefore, we can observe also that the shorter distance would be the period compared to the first result given.

Use the worked example above to help you solve this problem. A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 15.0 m/s. The point of release is h = 46.0 m above the ground.(a) How long does it take for the ball to hit the ground?
(b) Find the ball's speed at impact.
(c) Find the horizontal range of the ball.

Answers

Answer:

Explanation:

An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing 6.0 V for 0.65 ms. The resistance of the heart muscle between the pacemaker’s electrodes is 550 Ω. Find (a) the current that flows during a pulse, (b) the energy delivered in one pulse, and (c) the average power supplied by the pacemaker.