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Zolol [24]
2 years ago
6

One particular descent goes from 2100m to 1600m. Assuming work done against friction is 90% of the potential energy change of th

e cyclist and the cycle, what increase in speed in Km/h can a rider attain by the end of the
descent?
Physics
1 answer:
alisha [4.7K]2 years ago
7 0

Answer:

1/2 M V^2 = .1 M g H       where 10% of PE goes into KE

V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2

V = 31.1 m/s       increase in speed during descent

1 km / hr = 1000 m / 3600 sec = .278 m/s

V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr

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When a potential difference of 12 V is applied to a wire 6.7 m long and 0.31 mm in diameter the result is an electric current of
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The wire needs to be sauderwired to be connected back into place to get energy into column so it came function properly again!
7 0
3 years ago
Write a short summary paragraph that could go at the end of the text
UNO [17]

Answer:

The electro magnesium of the atoms cause the magnetic force to charge thus creating electricity. just kidding

Explanation:

If you read on the second paragraph it states that magnesium charged created electricity.

3 0
2 years ago
A dog has a mass of 20 kg. If the dog is pushed across the ice with a force of 40 N, what is its acceleration?
olasank [31]

Answer:

The acceleration is 2 m/s2.

Explanation:

We calculate the acceleration (a), with the data of mass (m) and force (F), through the formula:

F = m x a  ---> a= F/m

a = 40 N/20 kg                   <em>  1N= 1 kg x m/s2</em>

a= 40 kgx m/s2/ 20 kg

<em>a= 2 m/s2</em>

7 0
3 years ago
The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

6 0
3 years ago
What is the magnitude of your displacement when you follow directions that tell you to walk 225 m in one direction, make a 90° t
Gekata [30.6K]

Start by facing East. Your first displacement is the vector

<em>d</em>₁ = (225 m) <em>i</em>

Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,

<em>d</em>₂ = (350 m) <em>j</em>

Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,

<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )

<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>

The net displacement is

<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃

<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>

and its magnitude is

|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m

7 0
3 years ago
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