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suter [353]
3 years ago
11

A turtle takes 3.5 minutes to walk 18 m toward the south along a deserted highway. A truck driver stops and picks up the turtle.

The driver takes the turtle to a town 1.1 km to the north with an average speed of 12 m/s. What is the magnitude of the average velocity of the turtle for its entire journey?
Physics
1 answer:
Akimi4 [234]3 years ago
8 0

Answer:

3.59 m/s

Explanation:

The average velocity is defined as:

\frac{Total displacement}{Total time}

The turtle first walks 18m south, and then is taken 1,1Km (or 1100m) north. Thus, the total displacement  is 1082m north (1100m north - 18m south).

Now we have to calculate the total time, which will be equal to the sum of the time the turtle walked and the time it was taken by truck.

The walking time is 3.5 minutes. Since 1 minute = 60 seconds, then the walking time is 210 seconds.

To calculate the truck time we use the equation:

Time = \frac{Distance}{Speed}

Where the distance the truck travelled is 1100m and the speed of the truck is 12m/s.

Thus,

Truck time= \frac{1100m}{12m/s}= 91.67s

The total time is the sum of the walking time and the truck time.

Total time = 210s + 91.67s = 301.67s.

As mencioned previously, the average velocity is equal to total displacement/ total time, thus:

Average velocity = \frac{1082m}{301.67s} = 3.59 m/s North

Since the average velocity is a vector, it has a magnitude and a direction. In this case the magnitude is 3.59 m/s and the direction is north since the turtle's final displacement is north of where it started.

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Blababa [14]

Answer:

400 W/m^2 and 31℃

Explanation:

The output heat flux q"= 20 W/m^2 (geven)

The output heat flux from.the wall to the air by convection

q"conv = h(ts - t∞)

q"conv = 20(50-30) = 400 W/m^2

Therefor, this case is unsteady and the wall temperature changes with time till the energy balance exist.

ENERGY BALANCE

The input energy must be equal to the output energy for steady state condition. If not the state will be unstaidy or transient.

2. Its noticed that the output heat flux is not that the I put heat flux, therefore the wall tempers will be decreased till the output heat flux is reduced to the value of the given input heat flux

T steady = T∞ +q"/h

= 30 + 20/20 = 31℃

3 0
3 years ago
What is the Weight of Earth?<br><br><br><br><br> Don't SpAm​
Lena [83]

5.972 × 10^24 kg

it is the weight of earth

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6 0
2 years ago
Read 2 more answers
Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee
JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is 6.76\times10^{-12}\ J

The relativistic kinetic energy of a proton is 7.25\times10^{-12}\ m/s

Explanation:

Given that,

Mass of proton m=1.67\times10^{-27}\ kg

Speed v= 9.00\times10^{7}\ m/s

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Put the value into the formula

K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2

K.E=6.76\times10^{-12}\ J

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)

K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)

K.E=7.25\times10^{-12}\ m/s

Hence, The non-relativistic kinetic energy of a proton is 6.76\times10^{-12}\ J

The relativistic kinetic energy of a proton is 7.25\times10^{-12}\ m/s

7 0
3 years ago
A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a
Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

V_f= Final Velocity

Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
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Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

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So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
3 years ago
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