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algol13
3 years ago
5

The motorcycle travels with a constant speed v0 along the path that, for a short distance, takes the form of a sine curve. Deter

mine the x and y components of its velocity at any instant on the curve.
Physics
1 answer:
Alexandra [31]3 years ago
6 0

Let at any instant of time the speed is vo and the angle made by the bike with the horizontal is given

now we have

component of speed in x direction given as

v_x = v_0cos\theta

component of speed in y direction will be

v_y = v_0sin\theta

now from above two equations we can say that here

\theta = angle with the horizontal at any instant

and since here it is a sine curve so we know that

y = sin(x)

so we have slope of graph

tan\theta = \frac{dy}{dx} = cos(x)

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Assuming things about someone based on your experiences with similar people you have encountered is called
AysviL [449]

Answer:

presumptuous

Explanation:

it's what you call someone who assumes something

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3 years ago
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
2 years ago
Will give brainliest! how does an engineer use physical science?
pentagon [3]

Answer: gravity, circuits

Explanation:

3 0
2 years ago
The hydrogen stored inside a large weather balloon has a mass of 13.558 g. What is the volume of this balloon if the density of
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6 0
3 years ago
Read 2 more answers
A young kid of mass m = 36 kg is swinging on a swing. The length from the top of the swing set to the seat is L = 3.5 m. The boy
lara31 [8.8K]

Answer:

Explanation:

Given

mass of boy=36 kg

length of swing=3.5 m

Let T be the tension in the swing

At top point mg-T=\frac{mv^2}{r}

where v=velocity needed to complete circular path

Th-resold velocity is given by mg-0=\frac{mv^2}{r}

v=\sqrt{gr}=\sqrt{9.8\times 3.5}=5.85 m/s

So apparent weight of boy will be zero at top when it travels with a velocity of v=\sqrt{gr}

To get the velocity at bottom conserve energy at Top and bottom

At top E_T=mg\times 2L+\frac{mv^2}{2}

Energy at Bottom E_b=\frac{mv_0^2}{2}

Comparing two as energy is conserved

v_0^2=4gl+gl

v_0^2=5gL

v_0=\sqrt{5gL}=13.09 m/s

Apparent weight at bottom is given by

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6 0
2 years ago
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