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3 years ago

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Which of the following caught in or caught between hazard poses the greatest risk when working in excavations and trenches? A. B

eing pinned B. Cave-ins C. Equipment falling into trenches D. Severing of underground utilities

2 answers:

RSB [31]3 years ago

6 0

I believe the correct answer is b.) cave-ins

Vlada [557]3 years ago

4 0

A being pinned if a big bolder falls on you

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A car drives around a horizontal, circular track at constant speed. Consider the following three forces that act on the car: (1)

DiKsa [7]

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8 0

3 years ago

The area vector of a square loop of 5 turns of a conductor each with side length of 0.2 m carrying a current of 2 A is antiparal

Anon25 [30]

Answer:

$M= 0.4 Am^2$

Explanation:

From the question we are told that:

Number of turns $N=5$ $N=5$

Conductor each with side length $L=0.2m$

Current $I=2A$

Magnetic field $B=50.0T$

Generally the equation for the total magnetic moment M is mathematically given by

$M = current * area$

$M= I * A$

$M = 2 * (5* 0.2*0.2)$

$M = 2 * 0.2$

$M= 0.4 Am^2$

3 0

3 years ago

Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts

podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height =?

$y_{0}$ : is the initial height = 1 m

$v_{0_{y}$ : is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time

First, we need to find the time by using the following equation:

$t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s$

Now, the height is:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m$

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

$0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}$

$1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0$

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

$x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m$

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!

4 0

3 years ago

A car with mass m traveling at speed v has kinetic energy k. what is the kinetic energy of a second car that has the same mass m

vampirchik [111]

Kinetic energy, KE, is modeled by the formula $KE = \frac{1}{2}mv^2$ , where m is the mass in kg and v is the velocity in m/s.

In this scenario, mass and one-half are constant but the velocity changes.

You can see that by squaring twice the velocity, that is equal to four times the original KE. Therefore, the answer is 4k.

7 0

3 years ago

It took 3.5 hours for a train to travel the distance between two cities at a velocity of 120 km/h. How far did the train travel

Nadusha1986 [10]

Answer: 420 km

Explanation:

120 per hour, 3.5 hours

120 x 3.5 = 420

6 0

3 years ago