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BARSIC [14]
3 years ago
14

With what minimum speed must you toss a 110 g ball straight up to just touch the 11-m-high roof of the gymnasium if you release

the ball 1.3 m above the ground?
Physics
1 answer:
Sholpan [36]3 years ago
7 0

Answer:

 v = 13.79 m/s

Explanation:

given,

mass of ball = 110 g

height = 11 m

ball is released from = 1.3 m

minimum speed = ?

using conservation of energy

Potential energy is conserved in the form of kinetic energy

P E =KE

m g h= \dfrac{1}{2}mv^2

v^2 = 2 g h

v=\sqrt{2 g h}

v=\sqrt{2\times 9.8 \times (11-1.3)}

v=\sqrt{2\times 9.8 \times 9.7}

v=\sqrt{190.12}

       v = 13.79 m/s

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Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
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3 years ago
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Answer and Explanation:

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3 years ago
Hanna tosses a ball straight up with enough speed to rermain in the air for several seconds
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The mass luminosity relation L  M 3.5 describes the mathematical relationship between luminosity and mass for main sequence sta
ivanzaharov [21]

Answer:

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(b) <u>10 M</u>

Explanation:

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So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:

Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)

where,

L = Luminosity of Sun

M = mass of Sun

(a)

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Mass of Star = 2M

Therefore, eqn (1) implies that:

Luminosity of star/L = (2M/M)^3.5

Luminosity of Star = (2)^3.5 L

<u>Luminosity of Star =  11.3 L</u>

(b)

It is given that:

Luminosity of star = 3160 L

Therefore, eqn (1) implies that:

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taking ln on both sides:

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<u>Mass of Star = 10 M</u>

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4 years ago
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