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Margarita [4]
3 years ago
11

It is 6.00 km from your home to the physics lab. as part of your physical fitness program, you could run that distance at 10.0 k

m/hr (which uses up energy at the rate of 700 w ), or you could walk it leisurely at 3.00 km/hr (which uses energy at 290 w ).
Physics
1 answer:
kirill115 [55]3 years ago
7 0
<span>1.) It is 6.00km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10.0km/hr (which uses up energy at the rate of 700W ), or you could walk it leisurely at 3.00km/hr (which uses energy at 290 W). A.)Which choice would burn up more energy? running or walking? b.)How much energy (in joules) would it burn? c.)Why is it that the more intense exercise actually burns up less energy than the less intense one? Follow 2 answers Report Abuse Answers billrussell42 Best Answer: running, at 10 km/hour for 6 km is 6 km / 10 km/hour = 0.6 hour or 36 min energy used is 700 watts or 700 joules/s x 36 min x 60s/min = 1.512e6 joules or 1.5 MJ walking, at 3 km/hour for 6 km 6 km / 3 km/hour = 2 hour or 120 min energy used is 290 watts or 290 joules/s x 120 min x 60s/min = 1.872e6 joules or 1.8 MJ C) should be obvious PS, this has nothing to do with potential energy. billrussell42 · 5 years ago 0 Thumbs up 1 Thumbs down Report Abuse Comment Simon van Dijk I assume the watt consumption is per hour. Then running 6km at 10.0 km/h results in 700*6/10 = 420 w.h and walking in 290*6/3 = 580 w.h So walking would burn up more energy (kwh) b) 1 kilowatt hour = 3 600 000 joules so 420 wh = 0.42 kwh = 1.51.10^6 joule c) when you put more effort in making the distance your energy is used more efficient. Simon van Dijk · 5 years ago 0 Thumbs up 2 Thumbs down Report Abuse Comment</span>
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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
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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
3 years ago
To see if your results are reasonable, you can compare the final velocity of the stone as it falls down unwinding the wire from
pentagon [3]

Answer:

The velocity of the stone is 2.57 m/s.

Explanation:

Given that

Height = 0.337 m

We need to calculate the velocity of the stone

Using equation of motion

v^2-u^2=2gh

Where, v = velocity of stone

u = initial velocity

g = acceleration due to gravity

h = height

Put the value into the formula

v^2-0=2\times9.8\times 0.337

v=\sqrt{2\times9.8\times0.337}

v=2.57\ m/s

Hence, The velocity of the stone is 2.57 m/s.

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4 years ago
Describe three important ways we use the electromagnetic spectrum in our everyday lives.
OlgaM077 [116]

Answer:

We use X-rays to help the injured, Radiowaves to communicate or entertain, and Visible light to see.

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3 years ago
A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops
Ilya [14]

Answer:

(a) Compression work at the final state with a pressure of 1(MPa) is: 44.32(KJ), (b) Compression work at the final state with a pressure of 500(KPa): 110.37(KJ) and (c) temperaure of the final state in part b: T=151.83(°C).  

Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg andv_{2} =0.4v_{1} from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the T_{sat}=179.88^{0} C \leq T_{1} from the table 1 we get:v_{1} =0.30661(m^{3}/Kg). Now we have second conditions as: P_{2}=1(MPa), T_{2}=250^{0}C so from the same table we can see the state still superheated and we getv_{2}=0.23275(m^{3}/Kg), knowing that it's a isobaric process we can find the compression's work as:W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ) so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and:W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ) and the second process is isochoric:W_{2}=zero,nowW_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ). Finally to get the temperarure at the final state in part (b) we get:v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa) from table 2 (see attached) we comparev_{f} andv_{g} at the saturated water table and find the following:v_{f}=0.001093(m^{3}/Kg), so we know that the final state phase is a satured mixture and we get the temperature at the final state as:T_{2} =T_{sat} =151.83^{0}C.

3 0
4 years ago
A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the m
miskamm [114]

Answer:

4.9x10^-6T

Explanation:

See attached file

6 0
3 years ago
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