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Colt1911 [192]
3 years ago
14

Cara is swimming in a river. She can swim downstream up to a maximum speed of 66 meters per-minute. If she is swimming downstrea

m at half of her maximum speed, how far will she swim in 13 minutes?
Physics
2 answers:
makvit [3.9K]3 years ago
8 0
It would be 429 because in the question it said “ if she was swimming downstream of HALF of her maximum speed, how far will she swim for 13 minutes” so first you would divide 66 by 2 and get 33 then you would multiply 33 by 13 to get you’re answer.
VMariaS [17]3 years ago
7 0
429 meters
Half of 66 is 33 so that’s half of her maximum speed.
You then take 33 and times it by 13.
66/2=33
33x13=429
Hope this helps!
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At the surface, atmospheric pressure is 1.013 × 10^5 Pa. People can normally snorkel down to a depth of roughly one meter. What
natulia [17]

Answer:

1.01 × 10⁵ Pa  

Explanation:

At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.

We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.

Pressure at a depth is given by :

P=\rho gh

Where

\rho is the density of air, \rho=1.225\ kg/m^3

So,

P=1.225\times 9.8\times 1\\\\=12\ Pa

Total pressure, P = Atmospheric pressure + 12 Pa

= 1.013 × 10⁵ Pa + 12 Pa

= 1.01 × 10⁵ Pa

Hence, the total pressure is 1.01 × 10⁵ Pa.

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Mr. Jones's prescription calls for 1.04 tablets per day. Based on this information, how many tablets should Mr. Jones take per d
Katarina [22]

Answer:

C.) 1

Explanation:

1 is the closest to 1.04

Also you never want to go above your prescription amount

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Which factors affect friction between two solid surfaces?
a_sh-v [17]
1) newton of the surfaces in contact (the smoothness of the surfaces)

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A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 562 Hz tone, what is
tester [92]

To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,

Frequency = f = 562Hz

Speed of sound in air = v = 331m/s

The definition of wavelength is,

\lambda = \frac{v}{f}

Here,

v = Velocity

f = Frequency

Replacing,

\lambda = \frac{331m/s}{562Hz}

\lambda = 0.589m

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3 0
4 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.

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