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n200080 [17]
3 years ago
8

For each value below, enter the number correct to four decimal places. Suppose an arrow is shot upward on the moon with a veloci

ty of 52 m/s, then its height in meters after t t seconds is given by h ( t ) = 52 t − 0.83 t 2 h(t)=52t-0.83t2. Find the average velocity over the given time intervals.
Physics
1 answer:
kenny6666 [7]3 years ago
4 0

The time intervals are missing in the question.You can put any time in equation v(t) which I have solved

Answer:

v(t)=52-1.66t

Explanation:

Given data

Velocity=52 m/s

Height h(t)=52t-0.83t²

To find

Average velocity

Solution

As we know that velocity is first derivative of distance with respect to time

So

v(t)=\frac{dh(t)}{dt}\\ v(t)=\frac{d}{dt}[52t-0.83t^{2} ]\\ v(t)=52-1.66t

After one second the velocity is

v(t)=52-1.66t\\at\\t=1seconds\\v(1)=52-1.66(1)\\v(1)=56.33m/s

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Answer:

ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))

Explanation:

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3 years ago
A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

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3 years ago
Need help on science please!
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What does the model below represent?
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Answer:

D

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2 years ago
Mathphys :( im sorry i annoy you
Vitek1552 [10]

Answer:

4. 7.59276

Explanation:

Add up the x components:

Aₓ + Bₓ + Cₓ = 5 − 1.6 + 2.4 = 5.8

Add up the y components:

Aᵧ + Bᵧ + Cᵧ = -2.4 + 3.3 + 4 = 4.9

Use Pythagorean theorem to find the magnitude:

√(x² + y²)

√(5.8² + 4.9²)

√57.65

7.59276

3 0
3 years ago
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