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Elodia [21]
3 years ago
15

Why it is difficult to run fast in sand​

Physics
2 answers:
balu736 [363]3 years ago
8 0
Running on sand requires 1.6 times more energy spent than running on hard surface, so the force applied by our foot on sand is less.
prohojiy [21]3 years ago
5 0

Explanation:

The force exerted by our foot on the sand is less because time taken is large so it is difficult to run fast on sand.

hope it will help you

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If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is
expeople1 [14]

Answer:

We can use  2 g H = v2^2 - v1^2    or

v2^2 = 2 g H + v1^2

Since 88 ft/sec = 60mph   we have 30 mph = 44 ft/sec

The object will return with the same speed that it had initially so the object

starts out with a downward speed of 44 ft/sec

Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2

v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2

v2 = 110 ft/sec

8 0
3 years ago
i will give brainliest Which of the following structures are present in plant cells but absent in animal cells? Choose 1 answer:
Sonja [21]

Answer:

A. cell walls

Explanation:

Plants have cell walls but animals dont.

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3 years ago
A l'aide du document 1, écrire une synthèse sur la composition de l'atome en utilisant les mots
kirill115 [55]

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7 0
3 years ago
If the force on a hammer is 24 N and its mass is 1.6 kg, then the
Gnesinka [82]

Answer:

15 m/s²

Explanation:

F = ma

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6 0
3 years ago
A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can
Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

\mu mg = m\frac{v^2}{r}

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

\mu is the coefficient of friction between the tires and the road

m is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

v=85 mph \cdot \frac{1609}{3600}=38.0 m/s is the speed

And solving for \mu, we find the coefficient of friction required to keep the car in circular motion:

\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196

Learn more about circular motion:

brainly.com/question/2562955  

brainly.com/question/6372960  

#LearnwithBrainly

8 0
3 years ago
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