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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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jasenka [17]
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s = 1 275 000 000 km
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3 years ago
I need help witha worksheet over circuitsin physics could someone help??
garik1379 [7]
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3 years ago
You are trying to measure the mass of several different objects when you realize that there is a large wad of gum stuck to the u
nadya68 [22]

Answer:

accuracy

Explanation:

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6 0
2 years ago
Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0
3 years ago
How do mechanical waves to travel from one place to another?​
AnnZ [28]

Answer:

Mechanical waves require a medium in order to transport their energy from one location to another.

Sound waves are incapable of traveling through a vacuum. Slinky waves, water waves, stadium waves, and jump rope waves are other examples of mechanical waves; each requires some medium in order to exist.

4 0
3 years ago
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