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yarga [219]
3 years ago
5

Problem page a cyclist traveled 20 kilometers per hour faster than an in-line skater. in the time it took the cyclist to travel

80 kilometers, the skater had gone 30 kilometers. find the speed of the skater.
Physics
1 answer:
RideAnS [48]3 years ago
3 0
<span>Answer: skater x km/h cyclist 20 faster x + 20 km/h skater 30 km cyclist 80 km skater time = cyclist time t=d/r 30 / x = 80 /( x + 20 cross multiply 30 ( x + 20 ) = 80 x 30 x + 600 = 80 x 30 x - 80 x = -600 -50 x = -600 / -50 x = 12 km/h 12 km/h skater</span>
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A cyclist is moving at a speed of 15 m/s. If the combined mass of the bike and person is 100 kg,
MaRussiya [10]

Answer:

Explanation:

Momentum is equal to mass times velocity in kg and m/s, respectively. Therefore,

p = 100(15) so

p = 1500 \frac{kg*m}{s}

5 0
3 years ago
A cow wanders 30 m North, turns 22 degrees right of its original path, and wanders another 40 m. Find its total displacement.
scoundrel [369]

Answer:OB=58.3m

Explanation:

So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.

now take the starting point as a origin such that cow moves in x-y co-ordinate axis.

As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.

So the final displacement is the length of cow from the origin that is length OB.

now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]

now displacement of cow= length of OB

                                           = \sqrt{[37.084]^{2}+[44.984]^{2}  }

                                           =\sqrt{3398.78}

                                     OB =58.3

                                         

7 0
1 year ago
2. What is the power rating of an engine capable of lifting a 100 kg object 5 m vertically
Leya [2.2K]
Work=f.d
Work=100*50 = 500
Power = work/time = 500/4
=125 watt
7 0
3 years ago
A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle
emmasim [6.3K]

<u>Given data:</u>

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ       --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ  =>  R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>

<em>Pmin = W sin( α +Θ  )</em>

<em>          = W[ sin α.Cos Θ + cos α.sin Θ]</em>

<em>           = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>

<em>           = 460.9 N</em>

<em>Minimum push required to move the box up the ramp is 460.9 N</em>


7 0
3 years ago
I know how the voltage increasing/not enough/stays the same and how long the battery lasts with series and parallel connections.
kirill [66]

Answer:

It comes out the positive side of the battery and goes in to the negative side of the battery

Explanation:

There are already electrons in wires in a circuit before you add the battery. By adding the battery, you're giving the electrons the energy it needs to move along the circuit.

In a series circuit, the circuit is one continuous loop so there is only one path for the electrons to go - out of the positive side of the battery and around the circuit then goes back into the negative side of the battery.

However, with a parallel circuit, there are two or more ways the electrons can go so they take the path of least resistance. The electrons still go out the positive side of a battery but along the circuit, the electrons will go through the path of least resistance ( I tend to think of it like a net with holes in it - the lower the resistance the bigger the holes for the electrons to go through so more can fit in a set amount of time ) but the electrons still go out of the positive side and in through the negative

3 0
2 years ago
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