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katen-ka-za [31]
3 years ago
7

A body with the inertial

Physics
1 answer:
Andrews [41]3 years ago
3 0

Answer:

Explanation:

Hi there,

To get started, recall the kinematic equations from either a textbook, equation sheet, etc. Kinematic equations are used when acceleration is <em>constant,</em> as stated in the prompt.

Best way to use kinematic equations is to see which variable you are looking for, then which variable is unknown to you and is not needed for that equation.

a) average velocity

Takes the form of:

v_a_v_g=\frac{d_t_o_t_a_l}{t}=\frac{v+v_0}{2} this is the literal definition of average velocity; initial plus final divided by 2.

We know total displacement and total time elapsed, so we will use the middle form of the equation:

v_a_v_g=\frac{1640m}{40s}=41 \ m/s

b) the final velocity

We can still use the average velocity formula, as the other two equations that include final velocity have acceleration variable which is unknown as of now.

Solve for final velocity:

v=(2v_a_v_g)-v_o = 2(41 \ m/s) - (8 m/s) = 74 m/s\\ this makes sense, since a velocity later in time is higher than a velocity earlier in time. It is increasing with increasing time because of acceleration.

c) the acceleration

There are two equations that can be used to solve this, but we will use the less time-consuming one, but both produce same answer:

a = \frac{v-v_0}{t_t_o_t_a_l} = \frac{(74-8)m/s}{40s} =1.65 m/s^{2}

Notice, change in velocity over change in time, and acceleration is constant. When acceleration is constant, it models a linear function, and acc. is just slope!

Study well and persevere. If you liked this solution, hit Thanks or give a rating!

thanks,

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Modify how could you charge the electric circuit shown below to allow lightbulb a to stay lit even if lightbulb b is removed fro
shepuryov [24]
When a circuit is complete, or closed, electrons can flow from one end of a battery all the way around, through the wires, to the other end of the battery. Along its way, it will carry electrons to electrical objects that are connected to it – like the light bulb – and make them work!
5 0
3 years ago
Which statement about an atom is correct?
Alexus [3.1K]

Answer:

\boxed{ \sf{The \: electron \: has \: a \: negative \: charge \: and \: is \: found \: outside \: of \: the \: nucleus}}

Option A is correct

Explanation:

Let's check the options:

A: The electron has a negative charge and is found outside of the nucleus.

\mapsto{} Yeah! It's TRUE . An electron is a <u>negatively</u><u> </u><u>charged</u><u> </u><u>particle</u><u> </u><u>and</u><u> </u><u>is</u><u> </u><u>located</u><u> </u><u>outside</u><u> </u><u>the</u><u> </u><u>nucleus</u><u> </u><u>of</u><u> </u><u>an</u><u> </u><u>atom.</u>

B : The neutron has a negative charge and is found in the nucleus.

\mapsto{}No! It's FALSE . A neutron carries <u>no </u><u>charge</u>. i.e it is a neutral particle and found inside the nucleus.

C : The proton has no charge and is found in the nucleus.

\mapsto{}No! It's FALSE.A proton is a <u>positively</u><u> </u><u>charged</u><u> </u><u>particle</u> present inside nucleus of an atom.

D : The neutron has no charge and is found outside of the nucleus.

\mapsto{} I agree that the neutron has no charge. But it is found <u>inside</u> the nucleus not outside . So, this statement is FALSE .

Hence, we found our answer! :D

A. The electron has a negative charge and is found outside of the nucleus is the correct statement about an atom.

Hope I helped!

Best regards! :D

~\text{TheAnimeGirl}

7 0
3 years ago
A strip of copper 190 µm thick and 4.20 mm wide is placed in a uniform magnetic field of magnitude B = 0.78 T, that is perpendic
Veronika [31]

Answer:

V = 9.682 × 10^(-6) V

Explanation:

Given data

thick = 190 µm

wide = 4.20 mm

magnitude B = 0.78 T

current  i = 32 A

to find out

Calculate V

solution

we know v formula that is

V = magnitude× current / (no of charge carriers ×thickness × e

here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³

so put all value we get

V = magnitude× current / (no of charge carriers ×thickness × e

V = 0.78 × 32 / (8.47 x 10^28  × 190 × 1.602 x 10^(-19)

V = 9.682 × 10^(-6) V

3 0
3 years ago
The equation y = 5 Sin (3x - 4t), where
igomit [66]

Answer:

Frequency, f = 0.63 Hz

Period, T = 1.58 s

Speed of a wave, v = 1.34 m/s

Explanation:

The equation of a wave is given by :

y = 5 \sin (3x - 4t) ...(1)

y is in mm

x is in meters

t is in seconds

The general equation of a wave is given by :

y=A\sin(kx-\omega t) ...(2)

(i) Compare equation (1) and (2) we get :

k=3\\\\\omega=4

Since,

\omega=2\pi f\\\\f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{4}{2\pi}\\\\f=0.63\ Hz

(ii) Period of wave is :

T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.63}\\\\T=1.58\ s

(iii) Speed of a wave,

v=\dfrac{\omega}{k}\\\\v=\dfrac{4}{3}\\\\v=1.34\ m/s

4 0
3 years ago
Work out
Simora [160]

the father has to sit 0.5meter away from the kid because he is a 3/4 heavier that the kid

6 0
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