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andrey2020 [161]
3 years ago
15

Determine the maximum height and range of a projectile fired at a height of 6 feet above the ground with an initial velocity of

100 feet per second at an angle of 40 degrees above the horizontal.Maximum heightRange Question 20 options:a) 70.56 feet183.38 feet b) 92.75 feet310.59 feet c) 92.75 feet183.38 feet d) 70.56 feet314.74 feet e)
Physics
1 answer:
Molodets [167]3 years ago
8 0

Answer:

C is the correct answer

Explanation:

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The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if i
Sergio [31]

Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius=r=\frac{d}{2}=\frac{2.3}{2}=1.15 m

a.We have to find the angular velocity in radians per second if it spins=1200 rev/min

Frequency=\frac{1200}{60}=20 Hz

1 minute=60 seconds

Angular velocity=\omega=2\pi f

Angular velocity=2\times \frac{22}{7}\times 20=125.7 rad/s

b.We have to find the linear speed of its tip at this  angular velocity if the plane is stationary on the tarmac.

v=r\omega=1.15\times 125.7=144.56 m/s

c.Centripetal acceleration=\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2

Centripetal acceleration==\frac{18170.56\times g}{9.81}=1852.25 g m/s^2

7 0
3 years ago
A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55
Anit [1.1K]

Explanation:

We have,

Mass of a baseball is 0.147 kg

Initial velocity of the baseball is 44.5 m/s

The ball is moved in the opposite direction with a velocity of 55.5 m/s

It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Change in momentum,

\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s

Impulse = 14.7 kg-m/s

Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s

4 0
3 years ago
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, h
Alex73 [517]

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

5 0
3 years ago
Here is free pts!! Does anyone want to do some talking???
Stells [14]

Answer:

thxxx for pointsss

Explanation:

have a good day :)

3 0
3 years ago
Read 2 more answers
When thorium (90 protons) ejects a beta particle, the resulting nucleus has
Sav [38]

Answer:

none of the above

Explanation:

The actual answer is '91 protons'. In fact, the beta decay of the thorium-234 is the following:

{}_{90}Th \rightarrow {}_{-1}e+{}_{91}Pa

where inside the nucleus of Thorium (90 protons), a neutron turns into an electron (the beta particle) + a proton. Therefore, the resulting nucleus (which is Protoactinium) has a total of 90+1 = 91 protons.

So, the correct answer would be '91 protons'.

7 0
2 years ago
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