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lina2011 [118]
3 years ago
15

A man exerts more pressure while standing on one foot than by two foot​

Physics
2 answers:
Phoenix [80]3 years ago
7 0

Answer:While walking, one foot is in contact with the ground at a time. The surface area of one foot is less than two feet. Therefore, the pressure is more than the pressure exerted while standing, as the surface area of both feet is more.

Setler [38]3 years ago
6 0
Yes? Ok? That’s right??
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Two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2
lesantik [10]

Answer: Hello there!

We know this:

The distance between the cars at t= 0 is D.

car 2 has an initial velocity of v0 and no acceleration.

car 1 has no initial velocity and a acceleration of ax that starts at  t = 0

then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.

Because the cars are moving against each other, we want to se at what time t they meet, this is equivalent to see:  

position of car 1 + position of car 2 = D

and in this way we could ignore constants of integration :D

for the position of each car we integrate again:  

P1(t) = (1/2)ax*t^2 and P2(t) = v0t

v0t + (1/2)ax*t^2 = D

v0t + (1/2)ax*t^2  - D = 0

now we can solve it for t using the Bhaskara equation.

t = \frac{-v0 +\sqrt{v0^{2} + 4*(1/2)ax*D } }{2(1/2)ax} =\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}

that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).

Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:

v(\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}) = ax*\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax} = {-v0 +\sqrt{v0^{2} + 2ax*D }

where again, you need to replace the values of v0, D and ax.

7 0
3 years ago
Differentiate conductors and insulators with the help of suitable examples
andrew-mc [135]

Answer:

Any material that keeps energy such as electricity, heat, or cold from easily transferring through is an insulator

A conductor is a material which contains movable electric charges.

3 0
2 years ago
Read 2 more answers
If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the
PolarNik [594]

Answer:

a. v₁ = 16.2 m/s

b. μ = 0.251

Explanation:

Given:

θ = 15 ° , r = 100 m , v₂ = 15.0 km / h

a.

To determine v₁ to take a 100 m radius curve banked at 15 °

tan θ  = v₁² / r * g

v₁ = √ r * g * tan θ

v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s

b.

To determine μ friction needed for a frightened

v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg

v₂ = 4.2 m/s

fk = μ * m * g

a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²

a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²

F₁ = m * a₁  ,  F₂ = m * a₂

fk = F₁ - F₂   ⇒  μ * m * g = m * ( a₁ - a₂)

μ * g = a₁ - a₂   ⇒  μ = a₁ - a₂ / g

μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)

μ = 0.251

3 0
3 years ago
In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum
gulaghasi [49]

Answer:

Ratio of length will be \frac{L_2}{L_1}=1.108

Explanation:

We have given time period of the pendulum when length is L_1 is T_1=0.95sec

And when length is L_2 time period T_2=1sec

We know that time period is given by

T=2\pi \sqrt{\frac{L}{g}}

So 0.95=2\pi \sqrt{\frac{L_1}{g}}----eqn 1

And 1=2\pi \sqrt{\frac{L_2}{g}}-------eqn 2

Dividing eqn 2 by eqn 1

\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}

Squaring both side

\frac{L_2}{L_1}=1.108

8 0
3 years ago
What happens to a wave when it moves from one medium to another?
Nina [5.8K]
One side of the wave changes speed before the other side, causing the wave to move
3 0
3 years ago
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