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3 years ago

Name two energy transformations that occur as Adeline pedals her bicycle up a steep hill and then coasts down the other side.

2 answers:

6 0

Potential energy is first transformed into kinetic energy as she pedals, then gravitational as she coasts down the hill.

Mnenie [13.5K]3 years ago

3 0

Question

Name two energy transformations that occur as Adeline pedals her bicycle up a steep hill and then coasts down the other side.

Answer

Potential energy is first transformed into kinetic energy as she pedals, then gravitational as she coasts down the hill.

So it starts as potential engry sence they are staying still but then been they strt moveing by pedleing upnthe hill it turns/or changes in kentic energy sence we are gping up the when i go bake down i don't use as much force or have to push because the gravity is pulling me down thats when it turns into gravitational energy.

Hope this helps :)

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A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

BARSIC [14]

We are given that,

$\frac{dx}{dt} = 4ft/s$

We need to find $\frac{d\theta}{dt}$ when $x=8ft$

The equation that relates x and $\theta$ can be written as,

$\frac{x}{6} tan\theta$

$x = 6tan\theta$

Differentiating each side with respect to t, we get,

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}$

$\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}$

Replacing the value of the velocity

$\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2$

$\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta$

The value of $cos \theta$ could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

$cos\theta = \frac{6}{10}$

$\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2$

$\frac{d\theta}{dt} = \frac{24}{25}$

Search light is rotating at a rate of 0.96rad/s

4 0

2 years ago

PLEASE HURRYYY

babymother [125]

your answer is make up artist

4 0

3 years ago

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

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3 years ago

A concave lens can only form a. A. real image. B. reversed image. C. virtual image. D. magnified image.

Umnica [9.8K]

A concave lens can only form a virtual image. The correct option among all the options that are given in the question is the third option or option "C". Concave lenses are mostly thinner in the middle compared to its edges. I hope that this answer has come to your help.

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3 years ago

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Which is the best definition to describe a transfer of energy?

MissTica

Answer:

Explanation:

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2 years ago