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3 years ago

Why does it hurt more when you fall on concrete than on grass?

Physics

2 answers:

Alexus [3.1K]3 years ago

8 0

Answer:

Smaller forces are easier on your legs and you prefer to land on soft grass. Momentum/impulse explanation: Whether you land on concrete or soft grass, your change in momentum will be identical. A hard surface that brings you to a stop in 0.01 s requires a much larger force of 15,000 N.

Explanation:

IrinaVladis [17]3 years ago

5 0

Concrete is harder than the grass so that’s why

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2 kg

pochemuha

Answer:

2500 and kg its is very easy do with method

8 0

3 years ago

A sculpture is suspended in equilibrium by two cables, one from a wall and the other

aleksandrvk [35]

Answer:

$T_1=6655.295917 \approx 6655.3N$

Explanation:

From the question we are told that

Angle of cable 2 $\theta=37.0\textdegree$

Weight of sculpture $W=5000 N$

Generally the Tension from cable 2 $T_2$ is mathematically given by

$T_2sin37\textdegree=5000N$

$T_2=5000N/sin37\textdegree$

$T_2=8308.2N$

Generally the Tension from Cable 1 $T_1$ is mathematically given by

$T_1=T_2 cos37\textdegree$

$T_1=8308.2* cos 37\textdegree$

$T_1=6655.295917 \approx 6655.3N$

7 0

3 years ago

single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev

Goryan [66]

Answer:

The sound level of the 26 geese is $Z_{26}= 96.15 dB$

Explanation:

From the question we are told that

The sound level is $Z_1 = 81.0 \ dB$

The number of geese is $N = 26$

Generally the intensity level of sound is mathematically represented as

The intensity of sound level in dB for one goose is mathematically represented as

$Z_1 = 10 log [\frac{I}{I_O} ]$

Where I_o is the threshold level of intensity with value $I_o = 1*10^{-12} \ W/m^2$

$I$ is the intensity for one goose in $W/m^2$

For 26 geese the intensity would be

$I_{26} = 26 * I$

Then the intensity of 26 geese in dB is

$Z_{26} = 10 log[\frac{26 I }{I_o} ]$

$Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )$

$Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )$

From the law of logarithm we have that

$Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]$

$= 14.15 + 82$

$Z_{26}= 96.15 dB$

4 0

3 years ago

What is the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind?

salantis [7]

We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is

X=0m/s

From the question we are told that

mosquito flying 2 m/s

against a 2 m/s headwind

Generally

The speed over the ground is the Flight Speed minus resistance speed

Generally the equation for the speed over the ground is mathematically given as

X=Flight Speed-resistance speed

Therefore

X=2-2

X=0m/s

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8 0

2 years ago

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Yes this nees helppppppppppppppp

kozerog [31]

Answer:

lol

Explanation:

it was funny

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3 years ago