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bija089 [108]
3 years ago
13

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 shellE1  = \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:

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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
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