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Jlenok [28]
3 years ago
9

1. James drives 400 km in 5 hours to his grandmothers. What are the units for speed going to be?

Physics
1 answer:
Airida [17]3 years ago
6 0

Answer:

See the answer below

Explanation:

1. Speed is calculated as the ratio of distance and time. Hence, Jame's speed can be calculated as:

       400/5 km/hr = 80 km/hr

The unit for the speed would be km/hr. This can also be converted to m/s:

80 km = 80,000 m

1 hr = 3,600 s

80 km/hr = 80,000/3600 m/s = 22.22 m/s

2. Since James drove 400 km in 5 hours, the distance he drove is 400 km.

3. The time it took for James to get there is 5 hours.

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A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.40
Veseljchak [2.6K]

Answer:

53.895 m.

Explanation:

Using the equation of motion,

v² = u² + 2as .............. Equation 1

Where v = final velocity of the swan, u = initial velocity of the swan, a = acceleration of the swan, s = distance covered by the swan.

make s the subject of the equation,

s = (v² - u²)/2a----------- Equation 2

Given: v = 6.4 m/s, u = 0 m/s ( from rest)  a = 0.380 m/s².

Substitute into equation 2

s = (6.4²-0²)/(2×0.380)

s = 40.96/0.76

s = 53.895 m.

Hence the swan will travel 53.895 m before becoming airborne.

6 0
3 years ago
A rocket's acceleration is 6.0 m/s2. Assuming it starts at 0 m/s, how long will it take for the rocket to reach a velocity of 42
elena-s [515]
You said "<span>A rocket's acceleration is 6.0 m/s2.".

That just means that its speed increases by 6 m/s every second.
Whenever you look at it, its speed is 6 m/s faster than it was
one second earlier.

If it starts out with zero speed, then its speed is 6 m/s after 1 second,
12 m/s after 2 seconds, 18 m/s after 3 seconds . . . etc.

How long does it take to reach 42 m/s ?

Well, how many times does it have to go 6 m/s FASTER 
in order to build up to 42 m/s ?

That's just (42/6) = 7 times.

Writing it correctly, with the units and everything, it looks like this:


(42 m/s) / (6 m/s</span>²)

= (42/6)  (m/s) / (m/s²)

= (42/6)  (m/s · s²/m)

=  7 seconds
7 0
3 years ago
What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec
jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

E = \sigma T^{4}  (1)

Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

\lambda max = \frac{2.898x10^{-3} m. K}{T}   (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}  (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

T = \frac{2.898x10^{-3} m. K}{\lambda max}   (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

450 nm . \frac{1m}{1x10^{9} nm}  ⇒ 4.5x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}

T = 6440 K

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

700 nm . \frac{1m}{1x10^{9} nm}  ⇒ 7x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}

T = 4140 K

Comparison:

\frac{6440 K}{4140 K} = 1.55

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0
3 years ago
What is the difference between heat exhaustion and heat stroke?
balu736 [363]
I believe the answer is D, Heat exhaustion involves a lack of sweating, while heat stroke involves extreme sweating. Also just to add the on if heat exhaustion is left untreated then it could turn into a heat stroke.
6 0
3 years ago
Read 2 more answers
If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is
gavmur [86]
Kelvin is a base unit of temperature scale from SI that defines as zero degree Kelvin (absolute zero). The absolute zero is a hypothetical statement that all molecular movement stops because there is no transient of energy for the molecules to move. When converting temperature in degree Celsius to Kelvin, add 273. You are given 600K and you are asked to find it in degrees Celsius.  

T(K) = T(C) + 273
600 K = T(C) + 273
T(C) = 600 – 273
T(C) = 327 °C
<span>The answer is letter B.</span>
6 0
3 years ago
Read 2 more answers
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