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kkurt [141]
3 years ago
14

if a runners power is 400 watts as she runs, how much chemical energy does she convert into other forms in 10 minutes

Physics
1 answer:
AnnZ [28]3 years ago
3 0

Answer:

Energy converted = 240000\,Joules = 240\, kJoules

Explanation:

Recall that Power is the rate at which energy is transferred therefore defined by the mathematical formula: Power\,=\,\frac{Energy\,transferred}{time}

Since the information on the power of the runner is given, as well as the time the energy conversion takes place, we can then use this equation to find how much energy is been converted. Notice that we just need to change the given time *10 minutes) into the appropriate units  (seconds)to get the answer in SI units of energy (Joules). The conversion of 10 minutes into seconds is done by multiplying : 10 minutes * 60 seconds/minute = 600 seconds.

We use this then to find the energy converted by the runner:

Power\,=\,\frac{Energy\,transferred}{time}\\400 \,W = \frac{E}{600\,sec} \\400 \,W * 600\,sec=E\\E=240000\,Joules = 240\, kJoules

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The specific heat of a certain type of metal is 0.128 J/(g⋅∘C).0.128 J/(g⋅∘C). What is the final temperature if 305 J305 J of he
Makovka662 [10]

Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

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6 0
3 years ago
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to
Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0
3 years ago
Consider two identical insulated metal spheres, A and B. Sphere A initially has a charge of -6.0 units and sphere B initially ha
Oksi-84 [34.3K]

Answer:

<em>-2 units of charge</em>

Explanation:

charge on A = Qa = -6 units

charge on B = Qb = 2 units

if the spheres are brought in contact with each other, the resultant charge will be evenly distributed on the spheres when they are finally separated.

charge on each sphere will be = \frac{Qa + Qb}{2}

charge on each sphere =  \frac{-6 + 2}{2} = \frac{-4}{2} = <em>-2 units of charge</em>

8 0
3 years ago
In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step
satela [25.4K]

Answer:

W = 16.5 Kj

P = 49.9 Watt

E = 16471

Explanation:

m = 73.5kg

t = 5min 30sec = (5×60) + 30 = 330sec

each step = 16.6cm = 0.166m

h = 135×0.166 = 22.41 m

g = 10 m/s²

(i) W = F × s = W × h = mgh

W = 73.5×10×22.41 = 16471.35

W = 16.5 Kj

(ii) Power = workdone/time

P = 16471.35/330

P = 49.9 Watt

(iii) The energy burnt in this process = 16471

4 0
3 years ago
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