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dezoksy [38]
3 years ago
5

Given the picture below, which bullet was fired first?

Physics
2 answers:
mart [117]3 years ago
7 0
B was fired first and just trust me
olchik [2.2K]3 years ago
5 0

Answer:

b

Explanation:

i think

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“Why can I sometimes see the moon during the day?“Why can I sometimes see the moon during the day?
asambeis [7]

Answer:

its bc of the way earth spins

Explanation:

4 0
3 years ago
From a branch 35 m high, a 0.75 kg bird dives into a small fish tank containing
Bad White [126]

Answer:

ΔT = 1.22*10^-3 °C

Explanation:

First, you calculate the potential energy of the bird when it is at 35 m high. The potential energy is also the mechanical energy of the bird in this case.

U=mgh

m: mass of the bird = 0.75kg

g: gravitational constant = 9.8m/s^2

h: height = 35m

U=(0.75kg)(9.8m/s^2)(35m)=257.25\ J

All this energy is given to the water. You use the following formula in order to calculate the change in temperature:

Q=mc\Delta T

m: mass of the water = 50kg

c: specific heat of water = 4186 J/kg°C

Q is equal to U (potential energy of the bird) because the bird gives all its energy to water. By doing ΔT the subject of the formula you obtain:

\Delta T=\frac{Q}{mc}=\frac{257.25J}{(50kg)(4186J/kg°C)}=1.22*10^{-3}\ \°C

hence, the maximum rise in temperature is 0.00122 °C

7 0
3 years ago
How much energy is transferred to 1.327 grams of water when the
never [62]

Answer: option B is correct

Explanation:

8 0
3 years ago
A rescue pilot drops a survival kit while her plane is flying at an ultitude of 2500m with a forward velocity of 95m. If the air
never [62]

Answer:

Approximately 2.1\; \rm km, assuming that g = -9.8\; \rm m \cdot s^{-2}.

Explanation:

Let t denote the time required for the package to reach the ground. Let h(\text{initial}) and h(\text{final}) denote the initial and final height of this package.

\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial}).

For this package:

  • Initial height: h(\text{initial}) = 2500\; \rm m.
  • Final height: h(\text{final}) = 0\; \rm m (the package would be on the ground.)

Solve for t, the time required for the package to reach the ground after being released.

\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}.

\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}.

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at 95\; \rm m \cdot s^{-1}.) From calculations above, the package would travel forward at that speed for about 22.588\; \rm s. That corresponds to approximately:95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km.

Hence, the package would land approximately 2.1\; \rm km in front of where the plane released the package.

5 0
4 years ago
Can someone help with question.
Katarina [22]

Answer:

IDK

Explanation:

IDK

5 0
3 years ago
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