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larisa86 [58]
3 years ago
15

The position of a particle moving along the x axis is given by x = (21 + 22t – 6 t2) m, where t is in s. What is the average vel

ocity during the time interval t = 1 s to t = 4 s?
Physics
1 answer:
romanna [79]3 years ago
7 0

Answer:

12m/s

Explanation:

x = (21 + 22t – 6 t2) m,

We can express change bin velocity as

dx/dt= -12t + 22

But Velocity is changing

interval t = 1 s

t = 4 s

at t = 1, x = 21 + 22(1) -6(1)^2

=37

at t = 4, x = 21 + 22(4) -6(4)^2

=13

Distance travelled at interval t = 1 s

t = 4 s

X2 - X1=13- 37. = - 24m

Velocity= displacement/ time

= 24/(3-1)=12 m/s

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C, They change their shapes depending on their containers

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3 years ago
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A jet makes a landing traveling due east with a speed of 120 m/s .
vesna_86 [32]

Average acceleration over a time interval lasting \Delta t is

a_{\rm ave}=\dfrac{\Delta v}{\Delta t}

where \Delta v is the difference in the jet's final and initial velocities. It's coming to a rest, so

a_{\rm ave}=\dfrac{0-120\frac{\rm m}{\rm s}}{13.5\,\rm s}=-8.9\dfrac{\rm m}{\mathrm s^2}

so the average acceleration has magnitude 8.9 m/s^2 and is pointing West (the direction opposite the jet's movement, which should make sense because the jet is slowing down).

7 0
3 years ago
A 10.0-cm-long uniformly charged plastic rod is sealed inside a plastic bag. The net electric flux through the bag is 7.50 × 10
Rina8888 [55]

Answer:

66.375 x 10⁻⁶ C/m

Explanation:

Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as ;

∅ = Q / ε₀        -----------------(i)

Where;

∅ = 7.5 x 10⁵ Nm²/C

ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²

Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;

7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)

Solve for Q;

Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²

Q = 66.375 x 10⁻⁷ C

Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e

L = Q / l     ----------------------(ii)

Where;

Q = 66.375 x 10⁻⁷ C

l = length of the rod = 10.0cm = 0.1m

Substitute these values into equation (ii) as follows;

L = 66.375 x 10⁻⁷C / 0.1m

L = 66.375 x 10⁻⁶ C/m

Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.

3 0
3 years ago
Average velocity is different than average speed because calculating average velocity involves
Artemon [7]

Answer:

calculating displacement.

Explanation:

It's not true that displacement and distance would be the same always. Displacement is always smaller than or equal to distance as it is the smallest path between the initial and final point whereas distance is the measure of the total path covered.

8 0
2 years ago
What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?
MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

I = \frac{P}{A}

I = \frac{P}{4\pi r^2}

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

I = \frac{100}{4\pi (2.5)^2}

I = 1.2738W/m^2

The relation between intensity I and E_{max}

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Here,

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c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

E_{max}^2 = 2I\mu_0 c

E_{max}=\sqrt{2I\mu_0 c }

E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)

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Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

B_{max} = \frac{E_{max}}{c}

B_{max} = \frac{30.982 V /m}{3*10^8}

B_{max} = 1.03275 *10{-7} T

Therefore the maximum value of the magnetic field is B_{max} = 1.03275 *10{-7} T

3 0
3 years ago
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