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4 years ago

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there are different methods to responding to acute sports injuries depending upon whether the player is conscious or not. t or f

1 answer:

ASHA 777 [7]4 years ago

7 0

Your answer is true.

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A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

4 years ago

A race car travels 44.3 m/s around a banked (45° with the horizontal) circular (radius = 200 m) track. What is the magnitude of

djverab [1.8K]

The magnitude of the resultant force is given by the centripetal force, since the car is under a circular motion. So, we have:

$F_c=ma_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Where v is the linear speed and r the radius of the circular motion. Replacing this and solving:

$F=m\frac{v^2}{r}\\F=80kg\frac{(44.3\frac{m}{s})^2}{200m}\\F=785N*\frac{1kN}{1000N}\\F=0.785kN$

3 0

4 years ago

If the equation of an average velocity of a sport car was given by the equation V(t) = 3t^2 -6t +24. Determine its displacement

alisha [4.7K]

Answer:

72

Explanation:

The displacement of an object can be found from the velocity of the object by integrating the expression for the velocity.

In this problem, the velocity of the sport car is given by the expression

$v(t)=3t^2-6t+24$

In order to find the expression for the position of the car, we integrate this expression. We find:

$x(t)=\int v(t) dt=t^3-3t^2+24t+C$

where C is an arbitrary constant.

Here we want to find the displacement after 3 seconds. The position at t = 0 is

$x(0)=0^3-0+0+C=C$

While the position after t = 3 s is

$x(3)=3^3-3(3)^2+24(3)+C=72+C$

Therefore, the displacement of the car in 3 seconds is

$d=x(3)-x(0)=72+C-C=72$

7 0

3 years ago

How long will it take an object to hit the ground if dropped from 100 meters

mash [69]

Answer:

The answer depends on what object you are dropping. Are you dropping a balloon or a car? (I'm joking 'bout that one.) If the mass of the object is very little, then it might drop slower. If the mass is bigger, then it might drop faster.

Good luck!

Explanation:

8 0

4 years ago

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What would you have loved to press the pause button on so you could go deeper

kiruha [24]

Answer:

~Banana Fish~

3 0

3 years ago