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SVEN [57.7K]
3 years ago
14

A 350-g baseball is shot out of a small cannon with a velocity of 9.0 m/s. The baseball flies horizontally at a constant height

of 3.0 m. What is its angular momentum if its radius is 7.4 cm? (I know the answer is 9.6 kgm^2/s but I don't understand how to get it)
Physics
1 answer:
stira [4]3 years ago
4 0

Answer:

9.5 kg m^2/s

Explanation:

The angular momentum of an object is given by:

L=mvr

where

m is the mass of the object

v is its velocity

r is the distance of the object from axis of rotation

Here we have:

m = 350 g = 0.35 kg is the mass of the ball

v = 9.0 m/s is the velocity

r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)

Therefore, the angular momentum is:

L=(0.35)(9.0)(3.0)=9.5 kg m^2/s

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I need help in this small question:
Natali5045456 [20]

Answer:

1) The energy released during nuclear fission or fusion, especially when used to generate electricity is called nuclear energy.

2) It is not renewable because it is an element that has no way whatsoever to regenerate or replicate itself, nor gets created by any natural terrestrial means, neither makes itself available by arriving from outer space (like sunlight). There is a limited amount of it available on the Earth, and every bit you use is a bit you’ll never have available again (as Uranium atoms get destroyed by the fission process).

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5 0
3 years ago
In a redox reaction the substance that accepts electrons is said to be
netineya [11]
The question "<span>In a redox reaction the substance that accepts electrons is said to be?" is a bit vague. By definition, a "redox" or "reduction" reaction is one where classified by a gain of electrons. On the other hand, if it is a loss of electrons, then it is an oxidation reaction.</span>
8 0
3 years ago
Using Rayleigh's criterion, what is the smallest separation between two pointlike objects that a person could clearly resolve at
atroni [7]

 Answer:

y = 52.44 10⁻⁶  m

Explanation:

It is Rayleigh's principle that two points are resolved if the maximum of the diffraction pattern of one matches the minimum the diffraction pattern of the other

Based on this principle we must find the angle of the first minimum of the diffraction expression

         a sin θ= m λ

The first minimum occurs for m = 1

       sin θ =  λ  / a

Now let's use trigonometry the object is a distance L = 0.205 m

        tan θ = y / L

Since the angles are very small, let's approximate

        tan θ = sin θ/cos θ = sin  θ

        sin θ = y / L

We substitute in the diffraction equation

         y / L =  λ  / a

         y =  λ  L / a

Let's calculate

        y = 550 10⁻⁹ 0.205 / 2.15 10⁻³

        y = 52.44 10⁻⁶ m

7 0
4 years ago
A sphere of mass of 1.55 kg is accelerated upwards by a string to which the sphere is attached. Its speed increases from 2.81 m/
Mama L [17]

Answer:

The tension in the string is 16.24 N

Explanation:

Given;

mass of the sphere, m = 1.55 kg

initial velocity of the sphere, u = 2.81 m/s

final velocity of the sphere, v = 4.60 m/s

duration of change in the velocity, Δt = 2.64 s

The tension of the string is calculated as follows;

T =  ma  + mg\\\\T = m(a + g)\\\\where;\\\\a \ is \ upward \ acceleration \ of \ the \ sphere\\\\g \ is \ acceleration \ due \ to \ gravity =9.8 \ m/s^2\\\\a = \frac{\Delta V}{\Delta t} = \frac{v- u}{ t} = \frac{4.6 - 2.81 }{2.64} = 0.678 \ m/s^2

T = 1.55(0.678 + 9.8)

T = 1.55(10.478)

T = 16.24 N

Therefore, the tension in the string is 16.24 N

5 0
3 years ago
You are observing a spacecraft moving in a circular orbit of radius 100,000 km around a distant planet. You happen to be located
Natalija [7]

To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:

v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})

Here,

v_{o} =  Orbital Velocity

\lambda_{max} = Maximal Wavelength

\bar{\lambda}} = Average Wavelength

c = Speed of light

Replacing with our values we have that,

v_{o} = (3*10^5) (\frac{3.00036-3}{3})

<em>Note that the average signal is 3.000000m</em>

v_o = 36 km/s

Now using the definition about centripetal acceleration we have,

a_c = \frac{v^2}{r}

Here,

v = Orbit Velocity

r = Radius of Orbit

Replacing with our values,

a = \frac{(36km/s)^2}{100000km}

a= 0.01296km/s^2

a = 12.96m/s^2

Applying Newton's equation for acceleration due to gravity,

a =\frac{GM}{r^2}

Here,

G = Universal gravitational constant

M = Mass of the planet

r = Orbit

The acceleration due to gravity is the same as the previous centripetal acceleration by equilibrium, then rearranging to find the mass we have,

M = \frac{ar^2}{G}

M = \frac{(12.96)(100000000)^2}{ 6.67*10^{-11}}

M = 1.943028*10^{27}kg

Therefore the mass of the planet is 1.943028*10^{27}kg

7 0
3 years ago
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