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mestny [16]
3 years ago
5

Two groups of students made plans to help their community during a hurricane. The first group suggested training the community a

bout safe evacuation routes and nearby shelters. The second group suggested stocking food items in every house of the community. Why is training about safe evacuation routes a better method to help a community during a hurricane than stocking food items?
Select one:
a. It is easier to find evacuation routes than buy food items.
b. Evacuation routes are more environmentally friendly than food items.
c. It is less expensive to find evacuation routes than to buy food items.
d. Evacuation routes provide immediate safety from hurricanes unlike food items.
Physics
1 answer:
zysi [14]3 years ago
8 0
The correct option is D.
During hurricane, the most important thing is to get people out to safety in order to save their lives. Training the community about safety evacuation route is by far the better idea because it provide immediate safety from the natural hazard. The second idea is not a good one at all, because if food are stocked in every house in the community, the food will be destroyed alongside other things during the hurricane. 
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We equate the two equations, then we can solve for the time spent on the trip. Hope this answers the question. Have a nice day.
8 0
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How are science and technology the same
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Explanation:

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Sharon is driving on a straight road. She is driving north, and her speed is
sdas [7]

Answer:

b

Explanation:

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4 0
3 years ago
A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
A wheel of radius R = 0.80 m is pulled by a rope looped around a frictionless axle. The total mass of the wheel and axle assembl
iragen [17]

Answer:

\frac{K_T}{K_R}=\frac{600J}{188.63J}=3.18

Explanation:

To find the rotational kinetic energy you first calculate the angular acceleration by using the following formula:

\tau=I\alpha\\\\FR=I\alpha\\\\\alpha=\frac{FR}{I}

F: force applied

R: radius of the wheel

I: moment of inertia

\alpha=\frac{(120N)(0.80m)}{26.88kgm^2}=3.57\frac{rad}{s^2}

With this value you calculate the angular velocity:

\omega^2=\omega_o^2+2\alpha \theta\\

you calculate how many radians the wheel run in 5.0m

\theta=\frac{2\pi (0.8m)}{5.0m}=\approx1.00rad

\omega=\sqrt{2(3.57\frac{rad}{s^2})(1.00rad)}=2.67\frac{rad}{s}

Next, you use the formula for the rotational kinetic energy:

K_R=\frac{1}{2}I\omega^2=(26.88)(2.67)^2 J = 188.63J

For the transnational kinetic energy you use the following equation:

W=\Delta K_T  (net work equals the change in the kinetic energy).

By replacing the you obtain:

\Delta K_T=Fd=(120)(5.0)J=600J

Finally, the ratio between translational rotational kinetic energy is:

\frac{K_T}{K_R}=\frac{600J}{188.63J}=3.18

hence, translational kinetic energy is three times the rotational kinetic energy.

6 0
3 years ago
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