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expeople1 [14]
2 years ago
8

A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x

-axis.​
Physics
1 answer:
mezya [45]2 years ago
3 0

Answer:

-1.0778×10⁻¹⁰ N/C

Explanation:

Applying,

E = kq/r²................ equation 1

Where E = elctric field, q = charge, r = distance, k = coulomb's law

From the question,

Given: q = -3.0×10 C, r = 5.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values in equation 1

E = (-3.0×10)(8.98×10⁹)/5²

E = -1.0778×10⁻¹⁰ N/C

Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C

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A 1400.0 kg car crests a 3200.0 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and c
rjkz [21]

Answer:

a) v = 88.54 m/s

b) vf = 26.4 m/s

Explanation:

Given that;

m = 1400.0 kg

a)

by using the energy conservation

loss in potential energy is equal to gain in kinetic energy

mg × ( 3200-2800) = 1/2 ×m×v²

so

1400 × 9.8 × 400 = 0.5 × 1400 × v²

5488000 = 700v²

v² = 5488000 / 700

v² = 7840

v = √7840

v = 88.54 m/s

b)

Work done by all forces is equal to change in KE

W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)

we substitute

1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf²  -0 )

488000 = 700 vf²

vf² = 488000 / 700

vf² = 697.1428

vf = √697.1428

vf = 26.4 m/s

4 0
3 years ago
An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

Where,

\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

x = 11.3m

5 0
3 years ago
Three pendulums all have the same length and start from the same height. The first pendulum is very light and has a mass of 67 g
vovikov84 [41]

Answer:

All three pendulum will attain same velocity

Explanation:

All three pendulum will attain same velocity irrespective of their mass difference in isolated system (means where air drag are negligible) and at same length

As you know when velocity is calculated we can not take mass into account.  

3 0
3 years ago
A space station rotates, providing 1 g acceleration for its inhabitants. If it is
disa [49]

Answer:

Hype

Explanation:

Yyrru

4 0
3 years ago
How many electrons can exist on the following energy levels
alukav5142 [94]
2e level 1
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