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Dmitry_Shevchenko [17]
3 years ago
12

Two laws are described below:

Physics
1 answer:
sesenic [268]3 years ago
4 0
B, Both law a and b because they both in science terms are scientific laws
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Copper exists in nature as two isotopes. The atomic masses and relative abundance of these isotopes is given in the table. What
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7 0
3 years ago
A car travels 40 miles in 30 minutes.
lukranit [14]

Answer:

(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 m/s^2

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles =  40 x 1.60934

so 40 miles  =  64.3738 Km

similarly converting 30 minutes into hours

1 minute = \frac{1}{60}hours

30 minute = \frac{30}{60}hours

30 minute = \frac{1}{2}hours

Now

Average velocity = \frac{Speed}{time}

Substituting Values,

Average velocity = \frac{64.3738}{\frac[1}{2}}

Average velocity = 64.3738 \times 2

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting  velocity to m/s

128.74  \times \frac{5}{18}

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy  K= \frac{1}{2}mv^2

Substituting the values,

K= \frac{1}{2}1500(35.75)^2

K= \frac{1}{2}1500(1278.06)

K= \frac{1500 \times (1278.06)}{2}

K= \frac{1917093.75}{2}

K=958546.875 Joule---------------------------------------------(2)

(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

S= ut+\frac{1}{2}at^2

Substituting the known values,

s= ut+\frac{1}{2}at^2

s= (37.75)(15)+\frac{1}{2}a(15)^2

s=566.25+\frac{1}{2}a(225)

s=566.25+\frac{(225a)}{2}-------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

Using the formula

v=u +at

re arranging the formula we get,

a = \frac{v - u}{t}

Substituting the values

a = \frac{0 - 35.75}{15}

a = \frac{- 35.75}{15}

a= - 2.38 m/s^2----------------------------------------(4)

Now substituting 4 in 3 we get

s=566.25+\frac{(225( - 2.38)}{2}

s=566.25+\frac{-535.5}{2}

s=536.25-267.75

s=268.8m--------------------------------------------------------------(5)

4 0
3 years ago
What is the difference between direct current and alternating current?
IRINA_888 [86]
In a direct current, the electric charge, or current, only flows in one direction. In an alternating current, the electric charge changes periodically.  
5 0
3 years ago
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A 2-kg book falls off a shelf. It hits a student traveling 2 m/s. How much kinetic energy does the book have?
Gekata [30.6K]

4 J

1/2 mv^2 is ke formula

3 0
3 years ago
An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte
telo118 [61]

Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2

a. The angular acceleration is Torque divided by moments of inertia:

\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2

b. 5 revolution would be equals to 10\pi rad, or 31.4 rad. Since the engine just got started

\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5

\omega = \sqrt{1893.5} = 43.5 rad/s

c. Work done during the first 5 revolution would be torque times angular displacement:

W = T*\theta = 1930 * 31.4 = 60633 J

d. The time it takes to spin the first 5 revolutions is

t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s

The average power output is work per unit time

P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

P_i = T*\omega = 1930*43.5=83983 W or 84 kW

7 0
2 years ago
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