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3 years ago

10

A tractor trailer, an SUV and a sports car are all traveling at 55 mph and are the same distance from a road barrier. Which vehi

cle will hit the barrier with the greatest force of impact

1 answer:

monitta3 years ago

8 0

Whichever has the most mass so probably or trailer SUV

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More people end up in U.S. emergency rooms because of fall-related injuries than from any other cause. At what speed v would som

miss Akunina [59]

Answer:

$v_{f}=6.47m/s$

Explanation:

Given data

Distance d=7.00 ft= 7*(1/3.281) =2.1336m

Initial velocity vi=0m/s

To find

Final velocity

Solution

From Kinematic equation we know that:

$v_{f}^{2} =v_{i}^{2}+2gd\\v_{f}^{2}=0+2(9.81m/s^{2} ) (2.1336m)\\v_{f}^{2}=41.86\\v_{f}=\sqrt{41.86}\\v_{f}=6.47m/s$

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3 years ago

Identify the following as

bazaltina [42]

3. Kinetic energy
4. Potential energy
5. Kinetic energy because it’s moving towards the waterfall otherwise there wouldn’t be a waterfall.
6. Kinetic energy
7. Kinetic energy
8. Potential energy
9. Potential energy
10. Kinetic energy

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3 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

4 years ago

The atmosphere is held together by

Vera_Pavlovna [14]

Answer:

D. gravity

Explanation:

Gravity keeps the atmosphere from escaping into space.

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4 years ago

If you were to study the cephalic region of the

devlian [24]

Answer:

the standard way the body is positioned when using anatomical terminology ... invisible line that runs vertically through the center of the axial region.

Explanation:

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3 years ago