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____ [38]
3 years ago
11

200mg of uranium-238 has a half life of 4.5 billion years

Physics
1 answer:
Butoxors [25]3 years ago
8 0
How much 13.5 billons year
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What is a four chord song??
Agata [3.3K]

E, B, C# minor and A for the most songs

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4 years ago
The A note has a frequency of 440 Hz. What is the wavelength if the speed of sound is 343 m/s?
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3 years ago
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A horse travels 80 meters south in 20 seconds. What is its average velocity?
AURORKA [14]

Average velocity is a vector unit (i.e. includes magnitude <em>and </em>direction) calculated by working out distance ÷ time:

80 metres ÷ 20 seconds = 4 metres/seconds (m/s)

Therefore, your final answer is C. 4 m/s south.

7 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
During skidding of a vehicle, wheel moves ___________ rotation.
wariber [46]

Answer:

c it does not move as the tire stops and applys friction

Explanation:

6 0
3 years ago
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