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Salsk061 [2.6K]

2 years ago

9

a bus travelling on a straight road at 25m/s accelerates uniformly at 5m/s squared for 2 seconds. find its speed in kilometres p

er hour

1 answer:

notka56 [123]2 years ago

5 0

Answer:

126 kmh⁻¹

Explanation:

We can simply solve this by applying motion equations

where

v - final velocity

u - initial velocity

a-acceleration

t - time

v = u + at

<em>= 25 + 5×2 = 35 ms⁻¹</em>

<em> = (35/1000)×3600 = 126 kmh⁻¹</em>

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Why must the electric field be normal to the surface at every point of a charged conductor?

solong [7]

-- If the field were inclined to the surface, then it would have
some component parallel to the surface.

-- Then, since we're talking about a conductor, the charges
on the object would move in response to that component
of the field, until there was no longer any component of the
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3 0

2 years ago

How much heat is released as a 5.89 kg block of aluminum cools from 462 °C to 315 °C. The specific heat capacity of aluminum is

s2008m [1.1K]

Mass, m = 5890g
Change in temperature, θ = Final_temperature - Initial_temperature
= 315 - 462°C
= -147°C

Specific heat capacity of aluminum, c = 0.900 J/(g*K)
=mcθ
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Answer would be C.

5 0

3 years ago

Read 2 more answers

A force of 14 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i

enyata [817]

Answer:

$W = 19.8 J$

Explanation:

14 lb force is required to stretch the spring by 4 inch distance

So we have

$F = 14 lbf$

$F = 6.35 \times 9.8 N$

$F = 62.3 N$

stretch in the spring is given as

$x = 4 in = 0.1016 m$

now we will have

$F = kx$

$62.3 = k(0.1016)$

$k = 613.125 N/m$

Now we need to find the work to stretch it by x = 10 in = 0.254 m

so we have

$W = \frac{1}{2}kx^2$

$W = \frac{1}{2}(613.125)(0.254)^2$

$W = 19.8 J$

7 0

3 years ago

50POINTS! Find the orbital speed of a satellite in a circular orbit 1700km above the surface of the Earth. M_earth 5.97e24kg, r_

ivolga24 [154]

<h2>Answer: 7020.117 m/s</h2>

Explanation:

The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:

$V=\sqrt{G\frac{M}{R}}$ (1)

Where:

$G=6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}}$ is the gravity constant

$M_{Earth}=5.97{10}^{24}kg$ the mass of the massive body around which the satellite is orbiting, in this case, the Earth .

$R=r_{Earth}+h=8080000m$ the radius of the orbit (measured from the center of the planet to the satellite).

This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth $r_{Earth}$ and the altitude of the satellite above the Earth's surface $h$ .

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

$V=\sqrt{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})\frac{5.97{10}^{24}kg}{8080000m}}$

$V=7020.117\frac{m}{s}$

6 0

3 years ago

Read 2 more answers

Two long straight wires are parallel and carry current in the same direction. The currents are 8.0 and 12 A and the wires are se

Bess [88]

A magnetic field is a mathematical description of the magnetic influence of electric currents and magnetic materials. The magnetic field at any point is specified by two values, the direction and the magnitude; such that it is a vector field. Mathematically it is described as,

$B = \frac{\mu_0 I}{2\pi d}$

Here

$\mu_0$ = Permeability at free space constant

$I_{1,2}$ = Current at each object

d = Distance to the center point of the two object

Two magnetic field due to the current in the same directions then is,

$B = \frac{\mu_0 I_1}{2\pi d_1}+\frac{\mu_0 I_2}{2\pi d_2}$

Replacing,

$B = \frac{(4\pi *10^{-7})(8)}{2\pi (2*10^{-3})}+\frac{(4\pi *10^{-7})(12)}{2\pi (2*10^{-3})}$

$B = 8*10^{-4} +12*10^{-4}$

$B = 20*10^{-4}T$

Therefore the correct answer is E.

8 0

2 years ago