5 minute deadline List 2 academic and Scientific applications of radiation

Calculate the angle θ between the radius-vector of the point and the positive x axis (measured counterclockwise from the positiv

Y_Kistochka [10]

The point obviously is in the 3rs quadrant

So

စ= tan^-1( y/x)-180

စ= -89.7°

6 0

3 years ago

24. An elevator is moving vertically up with an acceleration a. The force exerted on the floor by a passenger of mass m is

Wewaii [24]

Given :-

An elevator is moving vertically up with an acceleration a.

To Find :-

The force exerted on the floor by a passenger of mass m .

Solution :-

As the man is in a accelerated frame that is non inertial frame , we would have to think of a pseudo force .

The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .

For the FBD refer to the attachment . From that ,

$\implies Weight_{apparent}= mg + ma$

Hence option d is correct choice .

I hope this helps .

3 0

2 years ago

g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest

ArbitrLikvidat [17]

Answer:

a. i. 3.43 m/s ii. 2.8 m/s

b. The thin-walled cylinder

Explanation:

a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

3 0

2 years ago

One object is thrown vertically upward with an initial velocity of 100 m/s and

gavmur [86]

We have that for the Question it can be said that The maximum height reached by the first object will be 100 times that of the other.

$(H_{max})_1=100*(H_{max})_2$

From the question we are told

One object is thrown vertically upward with an initial velocity of 100 m/s and another object with an initial velocity of 10 m/s. The maximum height reached by the first object will be

that of the other.

a. 10,000 times

b. none of these

c. 1000 times

d. 100 times

e. 10 times

Generally the equation for the velocity is mathematically given as

$v=\frac{d}{t}\\\\Where\\\\\frac{H_{max}_1}{H_{max}_2}=\frac{(V_1)^2}{(v_2)^2}\\\\\frac{H_{max}_1}{H_{max}_2}=\frac{10000}{(10}\\\\$

$(H_{max})_1=100*(H_{max})_2$

Therefore

The maximum height reached by the first object will be 100 times that of the other.

$(H_{max})_1=100*(H_{max})_2$

For more information on this visit

brainly.com/question/23379286

8 0

2 years ago

In comparing the Ptolemaic system to the Copernican system, neither is entirely correct since Ptolemy said the earth was at the

Nikitich [7]

In comparing the Ptolemaic system to the Copernican system, neither is entirely correct since Ptolemy said the earth was at the center and Copernicus said the planets had circular orbits. This is correct

5 0

3 years ago