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Alexus [3.1K]
3 years ago
9

Below is an incomplete pictorial representation of the situation described in this problem. Complete the sketch by drawing the e

lectric field due to each charge at point P. Make sure that all your vectors have the correct orientation.

Physics
2 answers:
Sidana [21]3 years ago
6 0

Answer:

Explanation:

This is it

r-ruslan [8.4K]3 years ago
6 0

Answer:

The direction of the electric field is the same as the connecting line between the charge and point P

Explanation:

The sketch is in the attached file below

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Momentum is conserved it can be transferred but not lost. true or false​
Anarel [89]

Answer: False

explanation: because Total momentum is conserved only if the collision can be considered isolated from the influence of net external forces

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If the speed is c=3.00•80 m/s and the frequency is 1030kHz=1030•30 Hz what’s the wavelength
Nana76 [90]

Answer:0.3meters

Explanation:

wavelength=velocity ➗ frequency

wavelength=300.80 ➗ 1030.80

Wavelength=0.3m

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3 years ago
Example 2.3 The moon is observed from
Natalija [7]

Explanation:

you can see this example to undersranding the question

8 0
3 years ago
Which part of the electromagnetic spectrum has a lower frequency than visible light?
miv72 [106K]

Answer:

infrared

Explanation:

Electromagnetic spectrum w.r.t. increasing frequency is

Radio waves

Microwaves

Infra red

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ultra violet

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3 0
3 years ago
A car with 2 × 10^3 kg moving at a speed of 10 m/s collides and sticks with car B of mass of 3 × 10^3 kg initially at rest. How
stepan [7]

Answer:

6 \times 10^4 \; \rm J.

Explanation:

KE lost = Total KE before Collision - Total KE after Collision.

For each car, the KE before collision can simply be found with the equation:

\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2, where

  • m is the mass of the car, and
  • v is the speed of the car.

The 2 \times 10^3\; \rm kg car would have an initial KE of:

\displaystyle \frac{1}{2} \times 2 \times 10^3 \times 10^2 = 10^5\; \rm J.

The 3 \times 10^3\; \rm kg car was initially not moving. Hence, its speed and kinetic energy would zero before the collision.

To find the velocity of the two cars after the collision, apply the conservation of momentum.

The momentum p of an object is equal to its mass m times its velocity v. In other words, p = m\cdot v.

Let the mass of the two cars be denoted as m_1 and m_2, and their initial speeds v_1 and v_2. Since the two cars are stuck to each other after the collision, their final speeds would be the same. Let that speed be denotes as v_3.

Initial momentum of the two-car system:

\begin{aligned}& m_1 \cdot v_1 + m_2 \cdot v_2 \\ &= 2 \times 10^3 \times 10 + 3 \times 10^3 \times 0 \\ &= 2 \times 10^4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}.

After the collision, both car would have a velocity of v_3 (since they were stuck to each other.) As a result, the final momentum of the two-car system would be:

m_1\cdot v_3 + m_2 \cdot v_3 = (m_1 + m_2)\, v_3.

Since momentum is conserved during the collision, the momentum of the system after the collision would also be 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}. That is: (m_1 + m_2) \, v_3 = 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}.

Solve for v_3:

\begin{aligned} v_3 &= \frac{(m_1 + m_2)\, v_3}{m_1 + m_2} \\ &= \frac{2 \times 10^4}{2 \times 10^3 + 3 \times 10^3} \\ &= \frac{2 \times 10^4}{5 \times 10^3} \\ &= 4 \; \rm m \cdot s^{-1}\end{aligned}.

Hence, the total kinetic energy after the collision would be:

\begin{aligned} &\frac{1}{2}\, m_1 \, v^2 + \frac{1}{2}\, m_2\, v^2 \\ &= \frac{1}{2}\, (m_1 + m_2)\, v^2 \\ &= \frac{1}{2} \times \left(2 \times 10^3 + 3 \times 10^3\right) \times 4^2 \\ &= 4 \times 10^4\; \rm J\end{aligned}.

The amount of kinetic energy lost during the collision would be:

\begin{aligned}&\text{KE After Collision} - \text{KE Before Collision} \\ &= 10^5 - 4 \times 10^4 \\&= 6\times 10^4\; \rm J \end{aligned}.

5 0
3 years ago
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