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olga nikolaevna [1]
2 years ago
14

Find the weight of an object with mass 80 kg on the moon ( g = 1.6 m/s^2)

Physics
1 answer:
SpyIntel [72]2 years ago
5 0

Answer:

80kg = 133 Newtons I'm pretty sure this is right.

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A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

6 0
3 years ago
A wave traveling in a string has a wavelength of 35 cm, an amplitude of 8. 4 cm, and a period of 1. 2 s. What is the speed of th
AlekseyPX

0.29 m/s (wave velocity = wavelength (lamda)/period (T) in metres)

35 / 1.2 = 29.16

29.16 ÷ 100 = 0.29

Wave velocity in string:

The properties of the medium affect the wave's velocity in a string. For instance, if a thin guitar string is vibrated while a thick rope is not, the guitar string's waves will move more quickly. As a result, the linear densities of the two strings affect the string's velocity. Linear density is defined as the mass per unit length.

Instead of the sinusoidal wave, a single symmetrical pulse is taken into consideration in order to comprehend how the linear mass density and tension will affect the wave's speed on the string.

Learn more about density here:

brainly.com/question/15164682

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3 0
1 year 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 length of the color attribute?
astraxan [27]

The length of the colour attribute is 6 characters.

<h3>What is a colour attribute?</h3>

Each colour has a distinct look based on three essential characteristics: hue, chroma (saturation), and value (lightness). It's critical to use all three of these properties when describing colour to appropriately identify it and distinguish it from others.

<h3>What is the use of colour attributes?</h3>

The HTML <font> color Attribute specifies the text color within the font element. Values of Attributes: colour name: It uses the colour name to set the text colour.

It should be noticed that colours are represented by the hex triplet.

Learn more about colour attributes here:

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6 0
1 year ago
What are some of the difficulties with measuring the volume of a gas? Select all that apply.
forsale [732]

Answer:

it is ---D because you can't measure gas and it's mass

HOPE THIS HELPS

6 0
2 years ago
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