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castortr0y [4]
3 years ago
5

Ok...... i need help

Physics
2 answers:
iragen [17]3 years ago
4 0
I can’t see the photo I even tapped on it I am so sorry we’ll have a nice day
ikadub [295]3 years ago
3 0

Answer:

Umm sorry i can't see it clearly.... I tried ;^;

Explanation:

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A bird is standing on an electric transmission line carrying 3000 A of current. A wire like this has about 3.0 x 10-5 22 of resi
Darya [45]

Answer:

13.5 x 10^-9 A

Explanation:

Yes

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Whats an astral projection
Dominik [7]

Answer:Esotericism

Explanation:

it’s something that’s in intentional out of body experience

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Item 19 Question 1 A Ferris wheel has a radius of 75 feet. You board a car at the bottom of the Ferris wheel, which is 10 feet a
sergey [27]

Answer:

y = 104.4 ft

Explanation:

As we know that we board in the car of ferris wheel at the bottom position

So we will have

final height of the car at angular displacement given as

y = y_o + R + R sin(270 - 255)

y = y_o + R + R sin 15

here we know that

y_o = 10 ft

R = 75 ft

so we have

y = 10 + 75 + 75 sin15

y = 104.4 ft

4 0
3 years ago
6. A ball rolled 125 m from the top of a hill to the bottom of a hill. How long
leva [86]
Time = distance / speed
T = 125/ 5
T = 25 meters per second
7 0
3 years ago
Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show
Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

m_2 = \frac{20}{3} kg

Part b)

Angular acceleration is given as

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = 176.6 N

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

m_1g r_1 = m_2 g r_2

20 g(0.5) = m_2 g(1.5)

10 = 1.5 m_2

m_2 = \frac{20}{3} kg

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

m_1g - T = m_1 a

also we have

T r_1 = I\alpha

now we have

m_1g = \frac{I a}{r_1^2} + m_1 a

a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}

a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}

a = 0.981 m/s^2

so angular acceleration is given as

\alpha = \frac{a}{r_1}

\alpha = \frac{0.981}{0.5}

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = \frac{I\alpha}{r_1}

T = \frac{45 (1.96)}{0.5}

T = 176.6 N

7 0
3 years ago
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