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evablogger [386]
3 years ago
6

A hot air balloon is ascending at a rate of 7.5m/s when a passenger drops a camera. If the camera is 25m above the ground when i

t is dropped, how much time does it take for the camera to reach the ground, and what is the velocity just before it lands?
Physics
1 answer:
Gemiola [76]3 years ago
5 0
The time it takes for an object to reach a certain level given the distance and the initial velocity can be calculated through the equation,

    d = V₀t + 0.5gt²

Where d is the distance, V₀ is the initial velocity, g is the acceleration due to gravity and t is time. Substituting the known values from the given above,

   25 = (-7.5)(t) + 0.5(9.8 m/s²)(t²)

The value of t from the equation 3.15s. Hence, it will take the object 3.15 seconds to reach the ground. 
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8 0
3 years ago
A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.
cupoosta [38]

Answer:

a) The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

\vec p = m\cdot \vec v (1)

Where:

\vec p - Vector momentum, measured in kilogram-meters per second.

m - Mass of the particle, measured in kilograms.

\vec v - Vector velocity, measured in meters per second.

If we know that m = 2.93\,kg and \vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right], then the momentum is:

\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]

\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]

The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}} (2)

\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right) (3)

Where:

\|\vec p\| - Magnitude of momentum, measured in kilogram-meters per second.

\theta - Direction of momentum, measured in sexagesimal degrees.

If we know that p_{x} = 8.731\,\frac{kg\cdot m}{s} and p_{y} = -11.661\,\frac{kg\cdot m}{s}, then the magnitude and direction of momentum are, respectively:

\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}

\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}

\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)

\theta \approx 306.823^{\circ}

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0
3 years ago
Calculate the density of a material that has a bat a mass of 52.457 g any volume of 13.5 cm³
3241004551 [841]
Answer:
d = 3.8857 g/cm^3

Explain:
Formula/: d = m/V

Hope this helps!
(a brainliest would be appreciated)
4 0
3 years ago
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