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4 years ago

You are caught on open water during a thunderstorm. what should you do?

1 answer:

6 0

Stay away from water. Try not to wash, shower, wash dishes, or have some other contact with water during a thunderstorm because lightning can travel through a building's plumbing.

Further Explanation:

Be aware

Check the climate conjecture before taking an interest in outside exercises. In the event that the estimate calls for tempests, defer your excursion or movement, or ensure sufficient safe house is promptly accessible.

Go inside

Keep in mind the expression, "When thunder thunders, go inside." Find a protected, encased asylum when you hear thunder. Safe havens incorporate homes, workplaces, strip malls, and hard-top vehicles with the windows moved up.

Hunch near the ground and isolated

On the off chance that you are gotten in an open zone, hunker down in a ball-like position (feet and knees together) with your head tucked and hands over your ears so you are down low with insignificant contact with the ground. Try not to rests. Lightning causes electric flows along the highest point of the ground that can be fatal more than 100 feet away. Hunching down is the best mix of being low and contacting the ground as meager as could be expected under the circumstances.

Discrete

On the off chance that you are in a gathering during a rainstorm, separate from one another. This will decrease the quantity of wounds if lightning strikes the

Answer Details:

Subject: Physics.

Level: High School

Key Words:

Be aware

Go inside

Hunch near the ground and isolated

Discrete

For further Evaluation:

brainly.com/question/6362387

brainly.com/question/12938965

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A 50 kg pitcher throws a baseball with a mass of 0.15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is

Andreas93 [3]

The velocity of the pitcher is <u>0.105 m/s</u> in a direction opposite to the velocity of the ball.

When no external force acts on a system, the total momentum of the system is conserved. The total initial momentum of the system is equal to the total final momentum of the system.

The pitcher and the ball are initially at rest, therefore, the total initial momentum of the system is zero.

Since no external forces act on the system comprising of pitcher and the ball, the total final momentum of the system is also equal to zero.

If the mass of the pitcher is mp and its speed is vp, the mass of the ball is mb and the ball's speed is vb, then the final momentum of the system of pitcher and the ball is given by,

$p=m_pv_p+m_bv_b=0$

Therefore,

$v_p=-\frac{m_b}{m_p} v_p$

Substituet 0.15 kg for mb, 50 kg for mp and 35 m/s for vb.

$v_p=-\frac{m_b}{m_p} v_p=-\frac{0.15 kg}{50 kg} (35m/s)=-0.105 m/s$

The pitcher has a velocity <u> 0.105 m/s</u> opposite to the direction of the velocity of the ball.

8 0

3 years ago

In a physics experiment, a ball is released from rest, and it falls toward the ground. The timer was not paying attention but es

tigry1 [53]

Answer:

(A) –14m/s

(B) –42.0m

Explanation:

The complete solution can be found in the attachment below.

This involves the knowledge of motion under the action of gravity.

Check below for the full solution to the problem.

4 0

4 years ago

9. A football punter attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in th

vichka [17]

Answer:

Angle is 55.52°

and Initial Speed is v=26.48 m/s

Explanation:

Given data

$x_{o}=0m\\ y_{o}=1.23m\\a_{oy}=a_{1y}=g=-9.8m/s^{2} \\x_{1}=67.0m\\y_{1}=0m\\t_{o}=0\\a_{ox}=m/s^{2} \\t_{1}=4.50s$

Applying the kinematics equations for motion with uniform acceleration in x and y direction

$x_{1}=x_{o}+v_{ox}t_{1}=67.0m\\0+4.50v_{o}Cos\alpha =67.0m\\v_{o}Cos\alpha =14.99\\v_{o}=14.99/Cos\alpha.....(1) \\and\\y_{1}=y_{o}+v_{oy}t_{1}+(1/2)a_{oy}t_{1}^{2} =0m\\ 1+4.50v_{o}Sin\alpha+(-9.8/2)(4.5)^{2}=0\\ v_{o}Sin\alpha=21.828.....(2)$

Put the value of v₀ from equation (1) to equation (2)

$\frac{14.99}{Cos\alpha }(Sin\alpha ) =21.828\\as\\tan\alpha =Sin\alpha /Cos\alpha \\So\\14.99tan\alpha =21.828\\tan\alpha =21.828/14.99\\\alpha =tan^{-1}(21.828/14.99) \\\alpha =55.52^{o}$

Put that angle in equation (1) or equation (2) to find the initial velocity

So from equation (1)

$v_{o}=(\frac{14.99}{Cos\alpha } ) \\v_{o}=(\frac{14.99}{Cos(55.52) } ) \\v_{o}=26.48m/s$

7 0

3 years ago

Question 10 of 34

labwork [276]

Julia walks from the park, which is six blocks east of her house, to the store, which is three blocks east of her house. Julia walks for 5 minutes. This walk's average speed will be 1.2 blocks per minute. Option B is correct.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points. Distance is the horizontal length covered by the body. While displacement is the shortest distance between the two points.

Displacement is a vector quantity .its unit is m.

The average velocity on this walk will be;

$\rm v_{avg}= \frac{d}{t} \\\\ \rm v_{avg}= \frac{6 \ block+ 3 \ block }{5 \ minute } \\\\ v_{avg}=1.4 \ block /min$

Hence option B is correct.

To learn more about displacement refer to the link; brainly.com/question/10919017

#SPJ1

3 0

2 years ago

How much work is done when an engine generates 400 Watts of power in 25 seconds?

IceJOKER [234]

Answer:

10000 J or 10 KJ

Explanation:

power = workdone/time taken

400 = workdone/25

workdone = 400 * 25

=10000 J

8 0

3 years ago

Read 2 more answers