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olga55 [171]
3 years ago
10

Carlos runs with velocity \vec{v}v →= (5.6 m/s, 29o north of east) for 10 minutes. How far to the north of his starting position

does Carlos end up? (Give your answer to the nearest meter)
Physics
1 answer:
romanna [79]3 years ago
7 0

Answer:

1629 metres

Explanation:

Given that Carlos runs with velocity v = (5.6 m/s, 29o north of east) for 10 minutes.

To the north, the velocity will be:

V = 5.6 × sin 29

V = 2.715 m/s

Convert the time to second

Time = 10 × 60 = 600

Using the formula below

Velocity = distance/ time

Substitute all the parameters into the formula

2.715 = distance/600

Distance = 2.715 × 600

distance = 1629 m

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HELP ASAP TIMED TEST
balu736 [363]

Answer:

<em>Correct choice: b 4H</em>

Explanation:

<u>Conservation of the mechanical energy</u>

The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):

E = U + K

The GPE is calculated as:

U = mgh

And the kinetic energy is:

\displaystyle K=\frac{1}{2}mv^2

Where:

m = mass of the object

g = gravitational acceleration

h = height of the object

v = speed at which the object moves

When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

U_1 = mgH

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

\displaystyle U_2=\frac{1}{2}mv^2

Since the energy is conserved, U1=U2

\displaystyle mgH=\frac{1}{2}mv^2    \qquad\qquad [1]

For the speed to be double, we need to drop the snowball from a height H', and:

\displaystyle mgH'=\frac{1}{2}m(2v)^2

Operating:

\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]

Dividing [2] by [1]

\displaystyle \frac{mgH'}{mgH}=\frac{4\frac{1}{2}m(v)^2}{\frac{1}{2}m(v)^2}

Simplifying:

\displaystyle \frac{H'}{H}=4

Thus:

H' = 4H

Correct choice: b 4H

4 0
3 years ago
Which physical activity would a global positioning system resource help you with?
PolarNik [594]

The physical activity that a global positioning system resource can help me with is : ( A )  Hiking

<h3>Function of Global positioning system ( GPS ) </h3>

GPS helps with the accurate measurement of physical activities  and factors such as location, time, elevation and so on. When hiking, the time, distance covered and location of the hiker  can be accurately measured with the use of the GPS.

Hence we can conclude that The physical activity that a global positioning system resource can help me with is  Hiking.

Learn more about GPS : brainly.com/question/9795929

#SPJ1

4 0
2 years ago
A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

4 0
2 years ago
Read 2 more answers
How do you find input energy with only the efficiency and output energy given?
iris [78.8K]

Answer:

power output/transformer efficiency multiply by 100%

Explanation:

derrive the formula from efficiency=power output/power input multiply by 100%

8 0
3 years ago
How do odd-shaped ceilings, decorative panels, draperies, and glass windows affect echo and noise?
vesna_86 [32]
I think they decrease echo and reduce noise, they do this by either absorbing vibrations or by scattering the sound so that echoes arrive at different times rather than reverberating as a standing wave. An echo is a reflection of a sound that arrives at the listener with a delay after the direct sound. The delay is usually proportional to the distance of the reflecting surface from the source and the listener.
6 0
3 years ago
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