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PilotLPTM [1.2K]
3 years ago
9

An "energy bar" contains 26 g of carbohydrates. For the steps and strategies involved in solving a similar problem, you may view

a Video Tutor Solution.
Part A If the energy bar was his only fuel, how far could a 68 kg person walk at 5.0 km/h?
Physics
1 answer:
masha68 [24]3 years ago
7 0

To solve this problem we will apply the definition of Power and Speed. In turn, we will consider that one gram of carbohydrate, according to numerous scientific studies, contributes around 17kJ of energy. Therefore, if this were true, the total energy of 26 grams would be

E = (26)(17000) = 4.42*10^5J

Power can be described as the amount of energy applied at a given time, that is,

P = \frac{E}{t} \rightarrow t = \frac{E}{P}

t = \frac{4.42*10^5}{380}

t = 1.16*10^3s

The speed is described as the distance traveled in a certain time, and its units in international system is m / s, converting and replacing we will have

v = 5km/h(\frac{1000m}{1km})(\frac{1h}{3600s})

v = 1.388m/s

Now,

v = \frac{d}{t} \rightarrow d = vt

The distance is,

d = vt

d = (1.388)(1.16*10^3)

d = 1610.08m

Therefore the distance walked is 1610.08m

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A 1.2 KG rubber ball is being thrown in the air if the ball is traveling at 2.0 M/S when it is 3.0 M off the ground what is the
Vitek1552 [10]

Answer:

37.7 J

Hope this helps! (see pictures)

6 0
3 years ago
A 7kg object is dropped on Earth,. Assuming it has not yet hit the ground, what is the velocity of the object after 2 seconds of
kirza4 [7]
The answer would be 9.8 meters/ sec^2.

7 0
3 years ago
Read 2 more answers
The blades in a blender rotate at a rate of 6100 rpm. When the motor is turned off during operation, the blades slow to rest in
MissTica

Answer:

<em>155.80rad/s</em>

Explanation:

Using the equation of motion to find the angular acceleration:

\omega_f = \omega_i + \alpha t

\omega_f is the final angular velocity in rad/s

\omega_i  is the initial angular velocity in rad/s

\alpha is the angular acceleration

t is the time taken

Given the following

\omega_f = 6100rpm

Time = 4.1secs

Convert the angular velocity to rad/s

1rpm = 0.10472rad/s

6100rpm = x

x = 6100 * 0.10472

x  = 638.792rad/s

\omega_f = 638.792rad/s\\

Get the angular acceleration:

Recall that:

\omega_f = \omega_i + \alpha t

638.792 = 0 + ∝(4.1)

4.1∝ = 638.792

∝ = 638.792/4.1

∝ = 155.80rad/s

<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>

5 0
2 years ago
A jogger travels a route that has two parts. The first is a displacement of 3 km due south, and the second involves a displaceme
spayn [35]

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

dy = \sqrt{d^2 - dx^2}

dy = \sqrt{3.85^2 - 3^2}  = 2.41 km

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

7 0
3 years ago
Need help asap please and thank you​
nordsb [41]
Distance = speed x time

distance = 116 x 10

distance = 1160 m
6 0
3 years ago
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