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3 years ago

What is the mass of 50 N object on earth

Physics

1 answer:

Aleksandr [31]3 years ago

6 0

The average gravity on Earth is:

$g=9.81\frac{m}{s^2}$

Since N represents Newtons and Newtons is a measurement of force we can use the following equation:

$F=ma$

Where F is force, m is mass, and a is acceleration. In this case the acceleration is the gravity acting on the object. Therefore, F=50N, a=g=9.81m/s^2 and m is to be determined. And so:

$F=ma\\\\F=mg\\\\50N=m(9.81\frac{m}{s^2} )\\\\\frac{50N}{9.81\frac{m}{s^2} } =m(\frac{9.81\frac{m}{s^2} }{9.81\frac{m}{s^2} } )\\5.0968kg=m\\\\m=5.1kg$

Therefore, the mass is approximately 5.1 kg.

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Tcecarenko [31]

So acceleration = (final velocity - initial velocity)/time

So (fv-iv)/t=a

(45-110)/4.5

Gives you (-130/9)km/h^2

You may have to convert the SI units so just follow my steps and change what must be changed

4 0

3 years ago

Read 2 more answers

A distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 n

maksim [4K]

Answer:

A) receding from the earth

B) $3.078x10^6m/s$

Explanation:

A) receding from the earth

The wavelength went from 434.1nm to 438.6nm, there was an increase in wavelength (also knowecn as redshift due to the doppler efft), this increase is due to the fact that the source that emits the radiation (the distant galaxy) is moving away and therefore the light waves it emits are "stretched", causing us to see a wavelength greater than the original.

B) $3.078x10^6m/s$

to calculate the relative speed we use the following formula:

$v_{rel}=c(1-\frac{\lambda_{1}}{\lambda_{2}} )$

where $c$ is the speed of light: $c=3x10^8m/s$

$\lambda_{1}$ is the wavelength emited by the source, and

$\lambda_{2}$ is the wavelength measured on earth.

we substitute all the values and do the calculations:

$v_{rel}=(3x10^8m/s)(1-\frac{434.1nm}{438.6nm} )\\\\v_{rel}=(3x10^8m/s)(1-0.98974)\\\\v_{rel}=(3x10^8m/s)(0.01026)\\\\v_{rel}=3.078x10^6m/s$

the relative speed is: $3.078x10^6m/s$

5 0

3 years ago

How could you increase electric force using each factor

NNADVOKAT [17]

Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. Electrical forces are extremely sensitive to distance.

5 0

4 years ago

A certain kind of glass has an index of refraction of 1.640 for blue light of wavelength 440 nm and an index of 1.595 for red li

DIA [1.3K]

Answer:

0.52°

Explanation:

refractive index for blue light, nb = 1.640

Refractive index for red light, nr = 1.595

Angle of incidence, i = 30°

Let the angle of refraction for blue light is rb and the angle of refraction for red light is rR.

By use of Snell's law for blue light

$n_{b}=\frac{Sin i}{Sin r_{b}}$

$1.64=\frac{Sin 30}{Sin r_{b}}$

$1.64=\frac{0.5}{Sin r_{b}}$

$r_{b}=17.75^{\circ}$

By use of Snell's law for red light

$n_{R}=\frac{Sin i}{Sin r_{R}}$

$1.595=\frac{Sin 30}{Sin r_{R}$

$1.595=\frac{0.5}{Sin r_{R}}$

$r_{R}=18.27^{\circ}$

The angle between the two beams, $\theta =r_{R}-r_{b}$

θ = 18.27° - 17.75°

θ = 0.52°

8 0

3 years ago

kogti [31]

Well, the figure seems to report that velocity is measured in m/s²... That label should say m/s. (Unless of course this is the graph of acceleration over time, but then the answer would probably be more complicated than the given choices.)

If the graph indeed shows velocity, and the unit is just a typo, then the displacement from A to D is equal to the area under the curve.

From A to B, the area is of a triangle with height 4 m/s and base 1 s, hence the area is 1/2 • (4 m/s) • (1 s) = 2 m.

From B to C, it's a rectangle with length 3 s and height 4 m/s, hence with area (3 s) • (4 m/s) = 12 m.

From C to D, it's a trapezoid with "height" 2 s and bases 4 m/s and 2 m/s, hence with area 1/2 • (4 m/s + 2 m/s) • (2 s) = 6 m.

The total displacement is then 2 m + 12 m + 6m = 20 m.

6 0

3 years ago