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vova2212 [387]
3 years ago
6

Someone pls answer this with steps

Physics
1 answer:
Ivanshal [37]3 years ago
8 0
No worry’s I am here to help
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Factors that affect pressure in liquids
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Two factors influence the pressure of fluids. They are the depth of the fluid and its density.
7 0
3 years ago
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A 3.0kg mass tied to a string
dem82 [27]

Answer:

\boxed{\sf Tension \ in \ the \ string \ (T) = 3 \ kN}

Given:

Mass (m) = 3.0 kg

Uniform speed (v) = 20 m/s

Length of string (r) = 40 cm = 0.4 m

To Find:

Tension in the string (T)

Explanation:

Tension (T) is the string will be equal to centripetal force (\sf F_c).

\boxed{ \bold{ T = F_c  =  \frac{m {v}^{2} }{r} }}

Substituting value of m, v & r in the equation:

\sf \implies T =  \frac{3 \times  {20}^{2} }{0.4}  \\  \\  \sf \implies T = \frac{3 \times 400}{0.4}  \\  \\  \sf \implies T =3 \times 1000 \\  \\  \sf \implies T =3000 \: N \\  \\ \sf \implies T =3 \: kN

\therefore

Tension in the string (T) = 3 kN

5 0
3 years ago
Two balls have their centers 2.0 m apart. One ball has a mass of m1 = 7.9 kg. The other has a mass of m2 = 6.1 kg. What is the g
maks197457 [2]

Answer:

3.036×10⁻¹⁰ N

Explanation:

From newton's law of universal gravitation,

F = Gm1m2/r² .............................. Equation 1

Where F = Gravitational force between the balls, m1 = mass of the first ball, m2 = mass of the second ball, r = distance between their centers.

G = gravitational constant

Given: m1 = 7.9 kg, m2 = 6.1 kg, r = 2.0 m, G = 6.67×10⁻¹¹ Nm²/C²

Substituting into equation 1

F = 6.67×10⁻¹¹×7.9×6.1/2²

F = 321.427×10⁻¹¹/4

F = 30.36×10⁻¹¹

F = 3.036×10⁻¹⁰ N

Hence the force between the balls = 3.036×10⁻¹⁰ N

8 0
3 years ago
What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m
masya89 [10]

Answer: 0.0058 m/s^{2}

Explanation:

Centripetal acceleration a_{C} is calculated by the following equation:

a_{C}=\frac{V^{2}}{r}

Where:

V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s is the Earth's orbital speed

r=1.5(10)^{11} m is the orbital radius

a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}

a_{C}=0.0058 m/s^{2}

4 0
3 years ago
What is matter in regards to energy?​
svp [43]

Answer:

Energy always involves. motion in some form. Matter is an object that that up space so it is. measured in units of space

Explanation:

8 0
3 years ago
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