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3 years ago

A block of mass 11.0 kg slides from rest down a frictionless 38.0° incline and is stopped by a strong spring with

Physics

1 answer:

ivanzaharov [21]3 years ago

8 0

Answer:

11.72 mm

Explanation:

The gravitational potential energy equals the potential energy of the spring hence

$PE_{gravitational}=PE_{spring}$

$mgh=0.5kx^{2}$ where m is the mass of object, g is the acceleration due to gravity, h is the height, k is the spring constant and x is the extension of the spring

$mgdsin\theta=0.5kx^{2}$ where \theta is the angle of inclination and d is the sliding distance

Making x the subject then

$x=\sqrt {\frac {2mgdsin\theta}{k}}$

Substituting the given values then

$x=\sqrt{\frac {2\times 11\times 9.81\times 3\times sin 38}{2.9\times 10^{4}}}= 0.117240716\approx 11.72 mm$

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Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

8 0

3 years ago

A magic medallion is suspended from a string inside a compartment of Hogwarts Express which is running straight westwards on hor

Serggg [28]

Answer:

a = 1 m/s² and

Explanation:

The first two parts can be seen in attachment

We use Newton's second law on each axis

Y axis

Ty - W = 0

Ty = w

X axis

Tx = m a

With trigonometry we find the components of tension

Sin θ = Ty / T

Ty = T sin θ

Cos θ = Tx / T

Tx = T cos θ

We calculate the acceleration with kinematics

Vf = Vo + a t

a = (Vf -Vo) / t

a = (20 -10) / 10

a = 1 m/s²

We substitute in Newton's equations

T Sin θ = mg

T cos θ = ma

We divide the two equations

Tan θ = g / a

θ = tan⁻¹ (g / a)

θ = tan⁻¹ (9.8 / 1)

θ = 84º

We see that in the expression of the angle the mass does not appear therefore you should not change the angle

4 0

3 years ago

The instantaneous emf resulting from magnetic induction equals the rate of change of flux is

katen-ka-za [31]

Lenz's Law: The polarity of the induced emf is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loop.

7 0

3 years ago

build three straight ramps on the stage one ramp should be nearly flat and another should be extremely steep the last ramp shoul

notsponge [240]

Answer:

hope this helps

Explanation:

3 0

2 years ago

a particle with a charge of 5.5 x 10^-8 c is 3.5 cm from a particle with a charge of -2.3 x10^-8 c. the potential energy of this

Yuri [45]

Answer:

-32.5 * 10^-5 J

Explanation:

The potential energy of this system of charges is;

Ue = kq1q2/r

Where;

k is the Coulumb's constant

q1 and q2 are the magnitudes of the charges

r is the distance of separation between the charges

Substituting values;

Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)

Ue= -32.5 * 10^-5 J

4 0

3 years ago