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Rus_ich [418]
3 years ago
11

A block of mass m 5.00 kg is

Physics
1 answer:
vekshin13 years ago
6 0
<span>a.
the force in the direction of movement
Fx = F x cosA*
Fx = 12 x cos25*
Fx = 10.88
m x a = 10.88
a = 10.88/5
a= 2.18 m/s^2
b.
when the block will be lifted off the floor and when the vertical component
Fy = mg
Fy = F x sin25*
Fy= 5 x 9.8
Fy= 115.94 N
c.
if Fx = ma
a = 115.94/5
a= 23.19 m/s^2
hope it helps
</span>
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In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-ca
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Answer:

A) He finds the same value of q / m for different materials , B)      y = ½ (q / m) E L² / v₀ₓ² , C) v = E / B , D)   B = 2.13 10⁻⁶ T, E) For the first part I have two off-center points., For the second part I can center one point but the other is off center

Therefore the third statement is correct

Explanation:

Part A

Thomson's experiments are the first proof that the atoms that until now were considered indivisible were constituted by different elements, in these experiments Thomson himself the ratio q / m of several cathodes and always found the same value, which allowed to establish that In atoms there are two types of particles, some of which are mobile and others are still.

When examining his statements the correct one is: He finds the same value of q / m for different materials

Part B

For this part let's use Newton's second law

        F = ma

        q E = m a

        a = q / m E

We use the kinematic relationship

          y = voy t - ½ to t²

          x = v₀ₓ t

The initial vertical velocity of electrons is zero

           y = ½ a (x / v₀ₓ)²

We replace

           y = ½ (q / m) E L² / v₀ₓ²

Part C

If there is no deflection, the electric and magnetic forces are the same and in the opposite direction

         Fm = Fe

         q v B = q E

          v = E / B

Part D

       

        We replace

        y = ½ (q / m) E L² / (E / B)²

         y = ½ (q / m) L² B² / E

If we do not want any deflection the magnetic field has to return the electrons the amount that they lower y = -4.12 cm

      -4.12 10⁻² = ½ q / m 0.12² B² / 1.1 10³

       -16.97 10⁻⁴ = 6.54 10⁻³ B² q / m

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      q / m = -1.758 10¹¹ C/ kg

      B = √ 0.259 1.758 10¹¹ = √ 4.55 10⁻¹²

      B = 2.13 10⁻⁶ T

Part E

As the charge that the two particles is different

For the first part I have two off-center points.

For the second part I can center one point but the other is off center

Therefore the third statement is correct

8 0
3 years ago
Consider a spherical planet of uniform density rho. The distance from the planet's center to its surface (i.e., the planet's rad
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Answer:

a) g(r) = 4\pi \cdot G \cdot \rho\cdot r, b) g = 4\pi \cdot G \cdot \rho\cdot r_{P}

Explanation:

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dg = G\cdot \frac{\rho \cdot dV}{r^{2}}

dg = G\cdot \frac{\rho \cdot dV}{r^{2}}

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g = 4\pi \cdot G \cdot \rho\cdot r_{P}

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Complete Question

The complete question is is shown on the first uploaded

Answer:

The elastic potential energy at point B is  PE_{elastic} = 50J

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Explanation:

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So     PE_{elastic} = 100 - 50 = 50J

     Now at point D

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