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serg [7]
3 years ago
11

A 5 kg toy is tied to a rope where the tension measures 150 N. What is the weight of the object?

Physics
1 answer:
elena-s [515]3 years ago
4 0
Formula from physics to get the answer.
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Approximately how fast is Jupiter orbiting the Sun? Approximately how fast is Jupiter orbiting the Sun? 10 km/skm/s a little les
Anna007 [38]

Answer:

The correct answer is

a little less than 15 km/s.

Explanation:

The distance between the sun and Jupiter varies by about 75 million km between the perihelion and the aphelion with an average distance of 778 million km from the sun for which it takes Jupiter about 12 years to complete its orbit round the sun giving it an orbital speed of about 13.07 km/s

The size of Jupiter is more than the twice the combined size of all the other planets, which is about 1.300 times the size of earth.

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4 years ago
What is the base unit to measure volume
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Liter
The Basic Unit for Volume
The liter, which is the basic unit of volume, was originally defined as the volume occupied by 1 kilogram of water, but today it's simply one-thousandth of a cubic meter
BRAINLY PLS
7 0
3 years ago
Upwelling _____.
Assoli18 [71]
- - A up welling is a solution appearing and moving towards the ocean surface meaning as, its going up,
Therefore, C <span>happens when water rises to replace moving surface water</span>
8 0
4 years ago
Read 2 more answers
The largest and the smallest balls used in the experiment are with diameter 9.52 mm, and 2.38 mm respectively. For a glycerin wi
svlad2 [7]

Answer

given,

largest diameter of  balls = 9.52 mm = 0.00476 m

                                 radius = 0.00476

smallest diameter of ball = 2.38 mm = 0.00238 m

                                 radius = 0.00119

viscosity = 1.5 Pa.s

density of the ball = 1.42 g/cm

F = 6\pi \eta r V_t

F = \dfrac{mv}{t}

F = \dfrac{\dfrac{4}{3}\pi\ r^3\times (\rho-\sigma) \times 0.99 V_T}{t}

6\pi \eta r V_t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{t}

t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{6\pi \eta r V_t}

t= \dfrac{0.22 r^2 (\rho-\sigma) }{\eta}

for small balls

t= \dfrac{0.22\times 0.00119^2 (1460-1300)}{1.5}

t = 0.033 ms

for larger ball

t= \dfrac{0.22\times 0.00476^2 (1460-1300)}{1.5}

t = 0.531 ms

6 0
4 years ago
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

4 0
4 years ago
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