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2 years ago

PLEASE HELPPPPP

Physics

1 answer:

Ganezh [65]2 years ago

5 0

The only graph that accurately depict the given motion is graph D.

The given parameters;

initial position of the man = 0

direction of the man's first displacement = backward
time of first motion, t₁ = 6 seconds
velocity of this first displacement = v₁
time without any motion (zero movement) = 6 seconds
direction of the second displacement = forward
velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

graph B, and

graph D.

The difference between graph B and D;

in graph B there is a uniform motion for 6 seconds

in graph D there is no motion for 6 seconds (this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds).

Thus, the only graph that accurately depict the given motion is graph D.

Learn more here: brainly.com/question/21095906

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A 62 kg boy and a 37 kg girl use an elastic rope while engaged in a tug-of-war on a friction-less icy surface. If the accelerati

Paraphin [41]

Answer:

The magnitude of the acceleration of the boy toward the girl is $1.31\ m/s^2$

Explanation:

It is given that,

Mass of the boy, $m_1=62\ kg$

Mass of the girl, $m_2=37\ kg$

The acceleration of the girl toward the boy is, $a_2=2.2\ m/s^2$

To find,

The acceleration of the boy toward the girl.

Solution,

Let $a_1$ is the magnitude of the acceleration of the boy toward the girl. We know that force acting on one object to other are equal in magnitude but opposite in direction. So,

$F_1=-F_2$

$m_1a_1=-m_2a_2$

$a_1=-\dfrac{m_2a_2}{m_1}$

$a_1=-\dfrac{37\times 2.2}{62}$

$a_1=-1.31\ m/s^2$

$|a_1|=1.31\ m/s^2$

So, the magnitude of the acceleration of the boy toward the girl is $1.31\ m/s^2$

7 0

3 years ago

A solenoidal coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long

Airida [17]

Answer:

$0.00027646\ T$

$2.33\times 10^{-5}\ H$

-0.04194 V

Explanation:

$N_2$ = Number of turns in outer solenoid = 330

$N_1$ = Number of turns in inner solenoid = 22

$I_1$ = Current in inner solenoid = 0.14 A

$\dfrac{dI_2}{dt}$ = Rate of change of current = 1800 A/s

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

r = Radius = 0.0115 m

Magnetic field is given by

$B=\mu_0\dfrac{N_2}{l}I\\\Rightarrow B=4\pi \times 10^{-7}\times \dfrac{330}{0.21}\times 0.14\\\Rightarrow B=0.00027646\ T$

The average magnetic flux through each turn of the inner solenoid is $0.00027646\ T$

Magnetic flux is given by

$\phi=BA\\\Rightarrow \phi=0.00027646\times \pi 0.0115^2\\\Rightarrow \phi=1.14862\times 10^{-7}\ wb$

Mutual inductance is given by

$M=\dfrac{N_1\phi}{I}\\\Rightarrow M=\dfrac{22\times 1.14862\times 10^{-7}}{0.14}\\\Rightarrow M=2.33\times 10^{-5}\ H$

The mutual inductance of the two solenoids is $2.33\times 10^{-5}\ H$

Induced emf is given by

$\epsilon=-M\dfrac{dI_2}{dt}\\\Rightarrow \epsilon=-2.33\times 10^{-5}\times 1800\\\Rightarrow \epsilon=-0.04194\ V$

The emf induced in the outer solenoid by the changing current inthe inner solenoid is -0.04194 V

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vovangra [49]

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