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3 years ago

12

A motorcycle is moving at a constant velocity of 15 meters/second. Then it starts to accelerate and reaches a velocity of 24 met

ers/second in 3 seconds. What’s the acceleration of the motorcycle over this time?

1 answer:

galben [10]3 years ago

5 0

3 meters per second

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Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which r

bagirrra123 [75]

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}$

where,

T = Radiated surface temperature

$\lambda_{max}$ = peak wavelength

(a)

here,

$\lambda_{max}$ = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}$

<u>T = 2987.6 k</u>

(b)

here,

$\lambda_{max}$ = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}$

<u>T = 19986.2 k</u>

6 0

3 years ago

An object is allowed to fall freely near the surface of a planet. The object has an acceleration due to gravity of 24 m/s2. How

Alborosie

Answer:

12 m

Explanation:

The object is in uniformly accelerated motion, so the distance covered can be found using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For this problem,

$g=24 m/s^2$

and

u = 0, since we are considering the first second of motion

So, substituting t = 1 s, we find

$s=0+\frac{1}{2}(24)(1)^2=12 m$

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3 years ago

Which is greater, the force exerted by the Earth on the Sun, or the force exerted by the Sun on the Earth? Why?

LekaFEV [45]

Answer:

There is no great force, the force exerted by the Earth on the Sun, and the force exerted by the Sun on the Earth are equal

Explanation:

By definition...

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3 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

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3 years ago

Identify indicators of a chemical reaction. Check all of the boxes that apply.

Misha Larkins [42]

Answer: 1,2,4

Explanation:

6 0

3 years ago