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-BARSIC- [3]
3 years ago
12

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 n frictional force. he pushes in a dire

ction 25.0? below the horizontal. what is the work done on the cart by the gravitational force?
Physics
1 answer:
alexandr1967 [171]3 years ago
5 0

Work is calculated as the product of Force, Distance, and angular motion. In this case, the work done by gravity is perpendicular to the motion of the cart, so θ = 90°

 and W=Fdcosθ

W=35.0 N x 20.0 m x cos90

W=0 J

This means that work done perpendicular to the direction of the motion is always zero.

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Si tu velocidad al caminar es 0,3 m/s. Determina el tiempo que demorarias en llegar al sol en horas (REALIZAR TRANSFORMACIÓN DE
Contact [7]

Answer:

138,516,546.9 horas.

Explanation:

Tenemos que usar la ecuación:

Velocidad = distancia/tiempo

Acá tenemos:

Velocidad = 0.3m/s

distancia = 149597870700 m

y queremos resolver la ecuación para el tiempo:

0.3m/s =  149597870700m/tiempo.

tiempo =  149597870700m/(0.3m/s) = 498,659,569,000 s

y sabemos que una hora tiene 3600 segundos, entonces si queremos transformar de segundos a horas tenemos:

498,659,569,000 s = (498,659,569,000/3600) h = 138,516,546.9 horas.

6 0
3 years ago
If a 10 kg ball and a 100 kg ball we’re dropped from a plane, which would hit the ground first?
meriva

Answer: The 100 kg ball

Explanation:

3 0
2 years ago
Is Mercury a heavier element than tin
Kryger [21]
Yes because mercury has more protons and electrons that tin. (30 more)
4 0
3 years ago
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t
Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

Charge of second Particle q_2=-8\ \mu C

distance between them d=4\ cm

k=9\times 10^{9}

magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}

E_1=27\times 10^7\ N/C (away from it)

Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

E_2=-18\times 10^7\ N/C(towards it)

E_{net}=E_1+E_2

E_{net}=9\times 10^7\ N/C(away from first charge)        

8 0
3 years ago
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