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Fed [463]
1 year ago
15

male lions and human sprinters can both accelerate at about 10.0 m/s2. if a typical lion weighs 170 kg and a typical sprinter we

ighs 75 kg, what is the difference in the force exerted by the ground during a race between these two species? (both the forward and normal forces should be calculated.)
Physics
1 answer:
Rasek [7]1 year ago
3 0

The difference in the force exerted by the ground during a race between these two species is 259.25

<h3>What is force?</h3>

Forces are influences that can change the movement of an object. A force can change the velocity of an object with mass, i.e. accelerate it. Forces can also be described intuitively by pushing or pulling. A force has both magnitude and direction and is a vector quantity.

Force exerted by the lion:

Normal force = mg

170 = m × 9.8

m = 17.34 kg

Forward force = ma

= 17.34 × 10.0

= 173.4 N

Total force = Normal force + Forward force

Total force = 170 + 173.4

= 343.4 N

Force exerted by the man:

Normal force = mg

75 = m × 9.8

m = 7.65 kg

Forward force = ma

= 7.65 × 10.0

= 76.5 N

Total force = Normal force + Forward force

Total force = 7.65 + 76.5

= 84.15 N

The difference in the forces exerted by these species is:

343.4 N - 76.5 N = 259.25 N

To know more about force visit:

brainly.com/question/19734564

#SPJ4

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You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
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Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

0.273 liters are needed to accomplish this task without boiling.

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How does superposition provide evidence for the evolution of earth?
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Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

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Axis y

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Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

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We seek tension for this case

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Let's look for the relationship between the tensions of the two cases

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   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

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