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SVEN [57.7K]
3 years ago
9

Describe a situation where an object has a changing velocity but constant speed.

Physics
1 answer:
ehidna [41]3 years ago
8 0
To summarize, an object moving in uniform circular motion is moving around the perimeter of the circle with a constant speed<span>. While the </span>speed<span> of the object is</span>constant<span>, its </span>velocity<span> is </span>changing<span>. </span>Velocity<span>, being a vector, has a </span>constant<span>magnitude </span>but<span> a </span>changing<span> direction.</span>
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Answer:

a. max potential energy

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The temperature of water in a beaker is 45°C. What does this measurement represent?
julia-pushkina [17]

Your answer would be C, with a quick lookup of this question online lol
8 0
3 years ago
Read 2 more answers
FIRST ONE TO ANSWER GETS A BRAINLY!!! PLZ HELP ME!!!
Mandarinka [93]

Answer:

The number of coils of the wire and the strength of the power source.

Explanation:

It is a direct relationship meaning when one increases so does the other. A stronger power source will induce a stronger magnetic force. The more coils the stronger the magnet.

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4 years ago
An AC voltage source and a resistor are connected in series to make up a simple AC circuit. If the source voltage is given by ΔV
ra1l [238]

Answer:

t = 5.48 × 10⁻³ s

Explanation:

Given:

ΔV = ΔVmax × sin(2πft)

frequency, f = 16.9Hz

thus,

ΔV = ΔVmax × sin(2π×16.9×ft)

Now,

Let 'R' be the resistance

Also according to the ohms law

i = V/R

where,

i = current

V = voltage

hence,

i=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}

also, given at time 't' the current in the circuit is 55.0% of the peak current

thus

i=\frac{55}{100}\times \frac{\Delta V_{max}}{R}=0.55\times \frac{\Delta V_{max}}{R}

thus,

0.55\times \frac{\Delta V_{max}}{R}=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}

or

0.55=sin(2\pi \times 16.9\times t)}

or

0.5823=(2\pi \times 16.9\times t)}

or

t = 5.48 × 10⁻³ s (Answer)

8 0
3 years ago
21.-Una esquiadora olímpica que baja a 25m/s por una pendiente a 20o encuentra una región de nieve húmeda de coeficiente de fric
jeyben [28]

Answer:

y = 12.82 m

Explanation:

We can solve this exercise using the energy work theorem

          W = ΔEm

friction force work is

          W = fr . s = fr s cos θ

the friction force opposes the movement, therefore the angle is 180º

           W = - fr s

we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular

           N -Wy = 0

           N = mg cos θ

the friction force remains

            fr = μ N

            fr = μ mg cos θ

             

work gives

           W = - μ mg s cos θ

initial energy

           Em₀ = ½ m v²

the final energy is zero, because it stops

we substitute

          - μ m g s cos θ = 0 - ½ m v²

          s = ½ v² / (μ g cos θ)

         

let's calculate

          s = ½ 20² / (0.55 9.8 cos 20)

          s = 39.49 m

this is the distance it travels along the plane, to find the vertical distance let's use trigonometry

            sin 20 = y / s

           y = s sin 20

           y = 37.49 sin 20

           y = 12.82 m

8 0
4 years ago
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