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1 year ago

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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.)

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:

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3 years ago

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Over 4 seconds, a car's momentum decreases by 1000 kg m/s how much force did it take to make this happen?

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Answer:

250N

Explanation:

Given parameters:

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2 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

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With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$

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3 years ago

The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo

a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

$t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s$

The tire angular velocity before stopping is:

$\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s$

Also its angular decceleration:

$\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2$

Using the following equation motion we can findout the angle it makes during the deceleration:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where $\omega$ = 0 m/s is the final angular velocity of the car when it stops, $\omega_0$ = 114rad/s is the initial angular velocity of the car $\alpha$ = 14.75 rad/s2 is the deceleration of the can, and $\Delta \theta$ is the angular distance traveled, which we care looking for:

$-114^2 = 2*(-14.75)*\Delta \theta$

$\Delta \theta = 440rad$ or 440/2π = 70 revelutions

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3 years ago