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Sergio [31]
2 years ago
3

It is 7.0 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h

(which uses up energy at the rate of 700 W), or you could walk it leisurely at 3.0 km/h (which uses energy at 290 W).
Required:
a. Which choice would burn up more energy, and how much energy (in joules) would it burn?
b. Why does the more intense exercise burn up less energy than the less intense exercise?
Physics
1 answer:
lord [1]2 years ago
8 0

Answer:

a. the second choice of leisurely walking at 3 km/h burn up more energy, which is equal to 2.436 MJ

b. Due to more time

Explanation:

a.

<u>FOR 10 km/h SPEED</u>:

First, we find time taken to reach physics lab:

S₁ = V₁t₁

t₁ = S₁/V₁

where,

t₁ = time = ?

S₁ = Distance = 7 km

V₁ = Speed = 10 km/h

Therefore,

t₁ = (7 km)/(10 km/h)

t₁ = (0.7 h)(3600 s/1 h) = 2520 s

Now, we calculate the energy burnt:

P₁ = E₁/t₁

E₁ = P₁ t₁

where,

E₁ = Energy Burnt = ?

P₁ = Power Used = 700 W

t₁ = 2520 s

Therefore,

E₁ = (700 W)(2520 s)

<u>E₁ = 1.764 MJ</u>

<u></u>

<u>FOR 3 km/h SPEED</u>:

First, we find time taken to reach physics lab:

S₂ = V₂t₂

t₂ = S₂/V₂

where,

t₂ = time = ?

S₂ = Distance = 7 km

V₂ = Speed = 3 km/h

Therefore,

t₂ = (7 km)/(3 km/h)

t₂ = (2.333 h)(3600 s/1 h) = 8400 s

Now, we calculate the energy burnt:

P₂ = E₂/t₂

E₂ = P₂ t₂

where,

E₂ = Energy Burnt = ?

P₂ = Power Used = 290 W

t₂ = 8400 s

Therefore,

E₂ = (290 W)(8400 s)

<u>E₂ = 2.436 MJ</u>

<u></u>

<u>Hence, the second choice of leisurely walking at 3 km/h burn up more energy, which is equal to 2.436 MJ</u>

b.

The more intense exercise burn up less energy than the less intense exercise, <u>because it is performed for a longer period of time.</u>

<u />

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is \frac{dT}{dt} = 1.016 ^oC/m

Explanation:

From the question we are told that

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    The time considered is  t =  10 \  seconds

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Generally the change in the bird temperature with time is mathematically represented as

      \frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 *  (5-x)] [-\frac{dx}{dt} ]

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From the given equation of velocity field

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So

\frac{dT}{dt} = -0.4[0.2t]  -0.6[-1.4] -0.2[2 *  (5-x)][ -[0.6x]]    

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Answer:

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(b) F =( 6.68*10¹¹⁴ i  + 7.27*10¹¹⁴ j  ) N

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To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

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v: velocity (m/s)

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Data

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B  =(1.40 T)i  

B  =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

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We apply the formula (1)

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1T= 1 N/ C*m/s

We apply the formula (1)

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F=\sqrt {42700^{2}+85000^{2}+(2*42700*85000)cos135^{o}}=62573.17217 N

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Answer:

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