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ruslelena [56]
2 years ago
9

The path of a particle is defined by y = 0.5x2. If the component of its velocity along the x-axis at x = 2 m is vx = 1 m/s, its

velocity component along the y-axis at this position is ____.
Physics
1 answer:
Dominik [7]2 years ago
8 0

Answer:

Velocity component along y-axis is 2 m/s.

Explanation:

Given that

y=0.2x^{2}--(1)\\\v_{y}=?\\

Differentiating (1) w.r.to t

\frac{dy}{dt}=\frac{d}{dt}(0.5x^{2})\\\\\frac{dy}{dt}=(0.5)(2x)\frac{dx}{dt}\\\\v_{y}=xv_{x}\\\\At\,x=2 ,\,v_{x}=1\,m/s\\\\v_{y}=2\,m/s

Velocity component along y-axis is 2 m/s.

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A semi-trailer is coasting downhill along a mountain highway when its brakes fail. The driver pulls onto a runaway truck ramp th
aleksklad [387]

<u>Answer:</u>

The velocity is 30.279 m/s

<u>Explanation</u>:

Consider the initial speed of the semi-trailer be v

Then, initial kinetic energy = \frac{1}{2} mv^2

According to question, the semi-trailer coast along a ramp, which is inclined at an angle of 170, and to a distance of 160m to stop

Change in vertical position =h=160 \times \sin 17^{\circ}-0= 46.779m

Final potential energy of semitrailer = mgh

Applying principle of conservation of energy,

\frac{1}{2} mv^2 = mgh

Solving for v, we get v^2 = 2gh = 2*9.8*46.779 = 916.8684

v^2 = 916.8684

v = 30.279 m/s

Therefore, the velocity is 30.279 m/s

6 0
2 years ago
A hobby rocket reaches a height of 72.3 meters and lands 111 meters from the launch point
faust18 [17]
Using the following formulas for projectile motion:

Height, H = ( Vo^2 * sin theta^2 )/g

Range, R = ( Vo^2 * sin 2*theta )/g

Rearranging in terms of Vo^2:

Vo^2 = gH / sin theta^2

Vo^2 = gR / sin 2*theta

Equating the two formulas to each other to solve for the angle theta:

gR / sin 2*theta = <span>gH / sin theta^2
</span>
Substituting the given values:

(9.8)(111) / sin 2*theta = (9.8)(72.3)<span> / sin theta^2
</span>angle = 52.36 degrees

Therefore, the angle of launch is approximately 52.36 degrees.
8 0
2 years ago
A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]). part a part complete find the amplitude
Pani-rosa [81]
The amplitude of a wave corresponds to its maximum oscillation of the wave itself. 
In our problem, the equation of the wave is
y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
y(x,t)=0.750 cm
and therefore this value corresponds to the amplitude of the wave.
4 0
2 years ago
Read 2 more answers
You skip north for 12 minutes to your best friend's house that is 1.5 kilometers away. What is your average velocity?
bagirrra123 [75]

Answer:

The average velocity is 7.5 km/h

Explanation:

Let's convert minutes to hours so our answer can be given in a common units of km/hour:

12 minutes = 12/60  hours = 0.2 hours

Now we estimate the average velocity calculating the distance travelled over the time it took:

1.5 / 0.2 km/h = 7.5 km/h

3 0
2 years ago
Dane is standing on the moon holding an 8 kilogram brick 2 metres above the ground. How much energy is in the brick's gravitatio
dimulka [17.4K]

The gravitational potential energy is 25.6 J

Explanation:

The gravitational potential energy (GPE) of an object is given by:

GPE = mgh

where

m is the mass of the object

g is the gravitational field strength

h is the height of the object above the ground

In this problem, we have

m = 8 kg is the mass of the brick

g = 1.6 N/kg is the gravitational field strength on the moon

h = 2 m is the height of the brick above the ground

Substituting,

GPE=(8)(1.6)(2)=25.6 J

Learn more about gravitational potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0
2 years ago
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