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OverLord2011 [107]
3 years ago
11

How to solve question #2?

Physics
1 answer:
Vera_Pavlovna [14]3 years ago
5 0
2a for example the first one,2sec. You know that every second it moves 3metres further. So 2x3=6 but you start at 0.50m so 6+0.50=6.5
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weight of Ali is 500andN.he is standing on the ground with an area of 0.025 m^2 area .we can find pressure under his feet. what
Inessa05 [86]

Answer:

20000 Pa

Explanation:

Pressure is defined as the force per unit area.

Mathematically : P =F/A    where F is force and A is area

Force = 500 N

Area= 0.025 m²

P= 500/0.025

P= 20000 Pa

8 0
2 years ago
A dog exerts a force of 30N to move a wagon 2m in 5s. What is the power of the dog
Hunter-Best [27]

Explanation:

power=f×v. recall= distances/ time

= f× d/t

= 30 × 2/5

=12watt

6 0
3 years ago
Which type of mirror produces images that are always upright and at the same distance from the mirror as the object is?
Alex73 [517]
Well, that would be a plane (flat) mirror 
<span>provided that </span>
<span>the mirror and the object are oriented parallel to each other</span>
7 0
3 years ago
Read 2 more answers
Jenny pushes a 40 N crate down the hall 2m. How much work did she do?
Shalnov [3]
Work done = force x distance = 40 x 2 = 80 Joules.
8 0
3 years ago
8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h
pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0
3 years ago
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