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dlinn [17]
3 years ago
7

A bullet is fired at a speed of 15 m/s from the edge of a cliff. The bullet strikes the ground 65 m from a height H. Find the fi

nal velocity of the bullet.
Physics
1 answer:
svet-max [94.6K]3 years ago
8 0

Answer:

v=38.73m/s

Explanation:

u=15m/s

v =?

s=65m

g=9.81

v²=u²+2×g×s

v²=15²+2×9.81×65=1500.3

v=38.73m/s

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Answer:

A

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3 years ago
A joule is an amount of energy, and a watt is a rate of using energy, defined as 1 W = 1 J / s. How many joules of energy are re
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4 0
3 years ago
Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th
Mariulka [41]

Answer:

a) 0.0288 grams

b) 2.6*10^{-10} J/kg

Explanation:

Given that:

A typical human  body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance  for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

a)

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

b)

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = \frac{energy \ absorbed }{mass \ of \ the \ body}

= \frac{1.10*10^6*1.6*10^{-14}}{80}

= 2.2*10^{-10} \ J/kg

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) =  1.2  *2.2*10^{-10} \ J/kg

Effective dose (Sv) = 2.6*10^{-10} J/kg

 

5 0
4 years ago
two objects are connected by a light string that passes over a frictionless pulley as in the figure below, where m1 A) a1=g<br>
Shtirlitz [24]
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5 0
3 years ago
A large piece of jewelry has a mass of 130.8 g. A graduated cylinder initially contains 47.7 mL water. When the jewelry is subme
Shkiper50 [21]

Answer: The density of this piece of jewelry is 8.90g/cm^3

Explanation:

To calculate the density, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Mass of piece of jewellery = 130.8 g

Density of piece of jewellery = ?

Volume of piece of jewellery =( 62.4-47.7 ) ml = 14.7 ml = 14.7cm^3   1cm^3=1ml

Putting values in above equation, we get:

\text{Density of piece of jewellery}=\frac{130.8g}{14.7cm^3}=8.90g/cm^3

Thus density of this piece of jewelry is 8.90g/cm^3

8 0
3 years ago
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