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jek_recluse [69]

3 years ago

12

A car traveling at 22 m/s comes to an abrupt halt in 0.1 second when it hits a tree. What is the deceleration in meters per seco

nd per second (i.e. m/s/s or m/s2)?

1 answer:

QveST [7]3 years ago

6 0

Answer:

220 m/s²

Explanation:

given,

initial speed of car= 22 m/s

final speed of car= 0 m/s

time taken by car to stop= 0.1 s

acceleration of car is equal to change in velocity per unit time

$a = \dfrac{\Delta v}{t}$

$a = \dfrac{v_f-v_i}{t}$

$a = \dfrac{0-22}{0.1}$

a = -220 m/s²

negative sign represent deceleration of the body

hence, deceleration of car is equal to a = 220 m/s²

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You have a circuit with a 50 Ω , a 100 Ω , and a 150 Ω - resistor connected in series. (a) Rank the current through them from hi

olga2289 [7]

Explanation:

(a) A circuit has 50 Ω, a 100Ω and a 150 Ω resistor are connected in series. We know that in series combination, current through each resistor is same. So, current through a 50 Ω, a 100Ω and a 150 Ω resistor is same.

(b) Ohm's law of given by :

V = I R

V is potential difference. As I is same, so, the resistor having highest resistance will have highest potential difference. So,

$V_3>V_2>V_1$ .

Hence, this is the required solution.

5 0

3 years ago

The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the w

Citrus2011 [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The magnetic field is $B_{net} = \frac{1}{4} * mT$

And the direction is $-\r k$

Explanation:

From the question we are told that

The magnetic field at the center is $B = 1mT$

Generally magnetic field is mathematically represented as

$B = \frac{\mu_o I}{2R}$

We are told that it is equal to 1mT

So

$B = \frac{\mu_o I}{2R} = 1mT$

From the first diagram we see that the effect of the current flowing in the circular loop is (i.e the magnetic field generated)

$\frac{\mu_o I}{2R} = 1mT$

This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

$B_1 = \frac{1}{2} \frac{\mu_o I}{2R}$

and for the larger semi-circular loop is

$B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction

So the net magnetic field would be

$B_{net} = B_1 - B_2$

$= \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

$=\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}$

$=\frac{\mu_o I}{8R}$

$= \frac{1}{4} \frac{\mu_o I}{2R}$

Recall $\frac{\mu_o I}{2R} = 1mT$

So

$B_{net} = \frac{1}{4} * mT$

Using the Right-hand rule we see that the direction is into the page which is $-k$

3 0

3 years ago

Two wheels having the same radius and mass rotate at the same angular velocity. One wheel is made with spokes so nearly all the

Mazyrski [523]

Answer:

C. The wheel with spokes has about twice the KE.

Explanation:

Given that

Mass , radius and the angular speed for both the wheels are same.

radius = r

Mass = m

Angular speed = ω

The angular kinetic energy KE given as

$KE=\dfrac{1}{2}I\omega ^2$

I=Moment of inertia for wheels

Wheel made of spokes

I₁ = m r²

Wheel like a disk

I₂ = 0.5 m r²

Now by comparing kinetic energy

$\dfrac{KE_1}{KE_2}=\dfrac{I_1}{I_2}$

$\dfrac{KE_1}{KE_2}=\dfrac{mr^2}{0.5mr^2}$

$\dfrac{KE_1}{KE_2}=2$

KE₁= 2 KE₂

Therefore answer is C.

5 0

3 years ago

The unit light-year is a measure of

Virty [35]

One light-year is the distance that light travels in vacuum
in one year. It's a unit of distance.

3 0

3 years ago

Read 2 more answers

A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a

olga nikolaevna [1]

Answer:

4 hoop, disk, sphere

Explanation:

Because

We are given data that

Hoop, disk, sphere have Same mass and radius

So let

And Initial angular velocity, = 0

The Force on each be F

And Time = t

Also let

Radius of each = r

So let's find the inertia shall we!!

I1 = m r² /2

= 0.5 mr² the his is for dis

I2 = m r² for hoop

And

Moment of inertia of sphere wiil be

I3 = (2/5) mr²

= 0.4 mr²

So

ωf = ωi + α t

= 0 + ( τ / I ) t

= ( F r / I ) t

So we can see that

ωf is inversely proportional to moment of inertia.

And so we take the

Order of I ( least to greatest ) :

I3 (sphere) , I1 (disk) , I2 (hoop) , ,

Order of ωf: ( least to greatest)

That of omega xf is the reverse of inertial so

hoop, disk, sphere

Option - 4

5 0

3 years ago