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2 years ago

10

What is the displacement of a student who walks 12 m North, 5 m West, 3 m South,

1 answer:

Contact [7]2 years ago

6 0

When you draw all 4 of those moves, you see that he winds up 9m North and 11m West of where he started.

So the magnitude of his displacement is

D = √(9² + 11²) m²

D = √(81 + 121) m²

D = √(202) m²

<em>D = 14.2 m</em>

The direction is arctan(-11/9) = <em>50.7° west of North</em>

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What is the source of the radioactivity?

Nataly_w [17]

The Earth itself is a source of terrestrial radiation. Radioactive materials including uranium, thorium, and radium exist naturally in soil and rock. Essentially all air contains radon<span> , which is responsible for most of the dose that Americans receive each year from natural background sources.</span>

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2 years ago

A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg · m2 and is rotating about a frictionless vertic

ryzh [129]

Answer:

0.2932 rad/s

Explanation:

r = Radius = 2 m

$I_i$ = Initial angular momentum = $275\ kgm^2$

$\omega_i$ = Initial angular velocity = 14 rev/min

$I_f$ = Final angular momentum

$\omega_f$ = Final angular velocity

Here the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s$

The final angular velocity is 0.2932 rad/s

7 0

2 years ago

A horizontal force of 675 N is needed to overcome the force of static friction between a level floor and a 300-kg crate. What is

White raven [17]

Explanation:

Below is an attachment containing the solution.

8 0

2 years ago

The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various

Schach [20]

Answer:

$q = -461532.5 \ C$

Explanation:

From the question we are told that

The electric filed is $E = 102 \ N/C$

Generally according to Gauss law

=> $E A = \frac{q}{\epsilon_o }$

Given that the electric field is pointing downward , the equation become

$- E A = \frac{q}{\epsilon_o }$

Here $q$ is the excess charge on the surface of the earth

$A$ is the surface area of the of the earth which is mathematically represented as

$A = 4\pi r^2$

Where r is the radius of the earth which has a value $r = 6.3781*10^6 m$

substituting values

$A = 4 * 3.142 * (6.3781*10^6 \ m)^2$

$A =5.1128 *10^{14} \ m^2$

So

$q = -E * A * \epsilon _o$

Here $\epsilon_o$ s the permitivity of free space with value

$\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}$

$q = -461532.5 \ C$

6 0

3 years ago

A fully loaded cart with a mass of 2200 kg starts from the top of a 12-meter hill on a roller coaster.

Salsk061 [2.6K]

Answer:

A. potential energy is 258720 Joule

Explanation:

A.Gravitational potential energy is: PE = m × g × h

velocity = 15.33 m/s when the car reaches the bottom of the hill.

where, m = mass

g = acceleration due to gravity

h = height from the bottom of hill.

The potential energy is : m×g×h

=(2200×9.8×12)

=258720 Joule

B. at the bottom of the hill, the potential energy is converted into kinetic energy so PE at top = KE at bottom

kinetic energy= $\frac{1}{2}$ ( $m*v^{2}$ )

where v = velocity

m= mass

therefore, v= $\sqrt\frac{2*K.E}{m} {}$

or, v= $\sqrt{\frac{2*258720}{2200} }$

or, v=15.33 m/s

7 0

2 years ago