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Lostsunrise [7]
3 years ago
12

Slow moving vehicles must display ___________ emblem at the rear to warn of their low speed.

Physics
2 answers:
Naddik [55]3 years ago
7 0

Answer:

Reflective emblem

Explanation:

A slow moving vehicle must display a reflective emblem at the rear to warn of their low speed. The sign of emblem is triangular in shape indicating that any sad incident have happened. Slow moving vehicles are generally prone to accident because of low visibility so they must display reflective emblem at rear.

KIM [24]3 years ago
6 0
<span>Slow moving vehicles must display reflective emblem at the rear to warn of their low speed.

The sign consists of a triangle and the primary purpose is to prevent any sad incident to happen, as slow moving vehicle (SMV) can cause serious accidents due to the less visibility at various times, therefore the sign must be visible enough.  </span>
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Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?
frosja888 [35]

Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

3 0
2 years ago
If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?
Law Incorporation [45]

a) 4.8 m/s² his average acceleration during the first 2.5 s

b) 15 m distance he cover during the first 2.5 seconds.

c) 31 m/s his speed as he finished the race.

<h3>(a) How to calculate the Average Acceleration ?</h3>

Average acceleration is calculated by using Newtons first law of motion

v = u + at

where

u is initial velocity

v is final velocity

a is constant acceleration

t is time

Given u = 0m/s , v = 12 m/s and t = 2.5 sec

Therefore average acceleration is given by

a=\frac{v}{t}

a=\frac{12}{2.5}

a = 4.8 m/sec²

b) Here we will use newtons second law of motion

s=ut+\frac{1}{2}at^{2}

On substituting value we get

s = 15m

c) Here we will use newtons third law of motion

v² = u² + 2as

Here u = 0 , a = 4.8 m/s² and s = 100 m

Therefore

v² = 960

v = 31 m/s

Disclaimer: the question was given incomplete in the portal. Here is the complete question.

Question: Usain Bolt's 100m sprint Runner set the world record for the100 meter sprint during the 2009World Championships in Berlin with a time of 9.58 s, reaching a top speed of 12 m/s in about 2.5 s.

a. What was his average acceleration during the first 2.5 s?

b. What distance did he cover during the first 2.5 seconds, assuming his acceleration was constant?

c. If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?

Learn more about Newtons law of motion here:

brainly.com/question/12525794

#SPJ4

5 0
2 years ago
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