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pochemuha
3 years ago
14

What does a constant velocity look like on a displacement vs time graph?

Physics
1 answer:
-Dominant- [34]3 years ago
8 0

Answer:

A line with slope equal to the velocity.

Explanation:

If one is in the presence of constant velocity, that means that at the quotient between displacement and time elapsed is a constant value, therefore one can write the following equation:

\frac{displacement}{time} =constant

therefore, solving for displacement we get:

displacement= constant \,*\, time

which if plotted with displacement (D) on the vertical axis  and time (t) on the horizontal axis, renders a line with slope equal to the constant value of the velocity (v):

D=v\,*\,t

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Which color in the visible spectrum has the highest frequency?<br> The Visible Light Spectrum
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It Is violet

Explanation:

It has the shortest wavelengths and the shorter the wavelengths the higher the frequency.

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What is the bump on the lateral side just proximal to the 5th metatarsal?
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4 years ago
A 1.1 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an
elixir [45]

Answer with Explanation:

Mass of block=1.1 kg

Th force applied on block is given by

F(x)=(2.4-x^2)\hat{i}N

Initial position of the block=x=0

Initial velocity of block=v_i=0

a.We have to find the kinetic energy of the block when it passes through x=2.0 m.

Initial kinetic energy=K_i=\frac{1}{2}mv^2_i=\frac{1}{2}(1.1)(0)=0

Work energy theorem:

K_f-K_i=W

Where K_f=Final kinetic energy

K_i=Initial kinetic energy

W=Total work done

Substitute the values then we get

K_f-0=\int_{0}^{2}F(x)dx

Because work done=Force\times displacement

K_f=\int_{0}^{2}(2.4-x^2)dx

K_f=[2.4x-\frac{x^3}{3}]^{2}_{0}

K_f=2.4(2)-\frac{8}{3}=2.13 J

Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J

b.Kinetic energy =K=2.4x-\frac{x^3}{3}

When the kinetic energy is maximum then \frac{dK}{dx}=0

\frac{d(2.4x-\frac{x^3}{3})}{dx}=0

2.4-x^2=0

x^2=2.4

x=\pm\sqrt{2.4}

\frac{d^2K}{dx^2}=-2x

Substitute x=\sqrt{2.4}

\frac{d^2K}{dx^2}=-2\sqrt{2.4}

Substitute x=-\sqrt{2.4}

\frac{d^2K}{dx^2}=2\sqrt{2.4}>0

Hence, the kinetic energy is maximum at x=\sqrt{2.4}

Again by work energy theorem , the  maximum kinetic energy of the block between x=0 and x=2.0 m is given by

K_f-0=\int_{0}^{\sqrt{2.4}}(2.4-x^2)dx

k_f=[2.4x-\frac{x^3}{3}]^{\sqrt{2.4}}_{0}

K_f=2.4(\sqrt{2.4})-\frac{(\sqrt{2.4})^3}{3}=2.48 J

Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J

3 0
4 years ago
The New Horizons probe that passed by Pluto during July 2015 is one of the fastest spacecraft ever assembled. It was moving at a
LenKa [72]

Answer:

2.91 x 10¹² sec

Explanation:

d = distance of nearest star, Proxima Centauri  = 4.3 ly = 4.3 x 9.46 x 10¹⁵ m

v = speed of new horizon probe = 14 km/hr = 14000 m/s

t = time taken for the new horizon probe to reach nearest star, Proxima Centauri = ?

Using the equation

d = v t

Inserting the values given

4.3 x 9.46 x 10¹⁵ = (14000) t

t = 2.91 x 10¹² sec

7 0
4 years ago
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