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Svetlanka [38]
3 years ago
14

X=(150^2•sin(2•42))/9.8

Physics
2 answers:
qwelly [4]3 years ago
8 0

Answer:

2283.341086

Explanation: Just plug the equation into the calculator as is

amid [387]3 years ago
3 0

Answer:

2283.3410863

Explanation:

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In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs
bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

V=\frac{I}{nAq}

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

\frac{I}{q} = 1.2*10^{18}

A= 1.3*10^{-8}m^2

n=6.3*10^{28} e/m^3

\omicron{O} = 1.2*10^{-4}(m/s)(N/c) Mobility

We can find the drift velocity replacing,

V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}

V= 1.465*10^-3m/s

The electric field is given by,

E= \frac{V}{\omicron{O}}

E=\frac{1.465*10^-3}{1.2*10^{-4}}

E=12.2V/m

7 0
3 years ago
HELP ASAP ILL GIVE BRAINLIST
OlgaM077 [116]

Answer:

Each of the joints represents a degree of freedom in the manipulator system and allows translation and rotary motion :) Hope this helps

4 0
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A parallel-plate capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is conne
icang [17]

Answer:

1) the capacitance of the capacitor increases. This is due to the induction of opposite charges on the two surfaces of the dielectric by the plate, this increased the charge in the field, from C =Q/v, it is seen that capacitance C will increase with increase in Q since v is constant.

2) the electric field intensity will also increase with increase in electric charges provided plate separation d remains constant.

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3 years ago
What is the farthest distance at which a typical "nearsighted" frog can see clearly in air?
Umnica [9.8K]

Answer: the correct option is D (17m).

Explanation: The farthest distance at which a typical "nearsighted" frog can see clearly in air is 17m.

7 0
3 years ago
A person's center of mass is very near the hips, at the top of the legs. Model a person as a particle of mass mm at the top of a
muminat

Answer:

\mathbf{v_{max} = \sqrt{gL}}

Explanation:

Considering an object that moving about in a circular path,  the equation for such centripetal force can be computed as:

\mathbf {F = \dfrac{mv^2}{2}}

The model for the person can be seen in the diagram attached below.

So, along the horizontal axis, the net force that is exerted on the person is:

mg cos \theta = \dfrac{mv^2}{L}

Dividing both sides by "m"; we have :

g cos \theta = \dfrac{v^2}{L}

Making "v" the subject of the formula: we have:

v^2 = g Lcos \theta

v=\sqrt{ gL cos \theta

So, when \theta = 0; the velocity is maximum

∴

v_{max} = \sqrt{gL \ cos \theta}

v_{max} = \sqrt{gL \ cos (0)}

v_{max} = \sqrt{gL \times 1}

\mathbf{v_{max} = \sqrt{gL}}

Hence; the maximum walking speed for the person  is \mathbf{v_{max} = \sqrt{gL}}

3 0
3 years ago
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