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alexandr402 [8]
3 years ago
14

Why is friction like applied force but different from gravity?

Physics
2 answers:
o-na [289]3 years ago
8 0
Unlike friction, gravity relies on a natural source. Gravity happens naturally, but friction happens when a force that opposes motion between any surfaces that are touching.
kirill115 [55]3 years ago
3 0

Answer:

Friction is when a force is applied or done by weight dragging onto something.

Explanation:

Gravity is when an object is getting pulled toward the center of what is attracting it. And applied force is when someone/sommething is applying force.

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Lean your shoulders back and your waist forwards. Use your arms as a counter weight.
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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
Brrunno [24]

Answer:

The explorer should travel to reach base camp to 5.02 Km at 4.28° south of due west.

Explanation:

Using trigonometric function like Sen(Ф), Cos(Ф) and Tan(Ф) we can get distance and direction that the explorer should travel to reach base camp. When we discompound the vector X = 7.8*Cos(50) = 5.01 Km y y = 7.8 * Sen (50) - 5.6 = 5.975 - 5.6 (Km) = 0.375 (Km) so that Tan (\alpha ) = \frac{0.375}{5.01} = 4.28; \alpha = Arctang (\frac{0.375}{5.01}) = 4.28 (degree) to get how far we use Pythagorean theorem so R^{2} = x^{2}+y^{2} so that R=\sqrt{0.375^{2}+5.01^{2} } =5.02 (Km)

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3 years ago
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The force against a solid object moving through a gas or a liquid
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This drag force is always opposite to the object's motion, and unlike friction between solid surfaces, the drag force increases as the object moves faster.
3 0
3 years ago
A 500 N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area
Ghella [55]

Answer:

d. 10000N

Explanation:

When a force (F_1) is exerted on the smaller area piston (A_1), the pressure that originates therein is transmitted to the larger area piston(A_2). According to Pascal's principle the pressure on the smallest piston (P_1=\frac{F_1}{A_1}) will be equal to the pressure on the largest piston (P_2=\frac{F_2}{A_2}):

\frac{F_1}{A_1}=\frac{F_2}{A_2}\\F_2=F_1\frac{A_2}{A_1}\\F_2=500N\frac{40cm}{2cm}\\F_2=10000N

7 0
3 years ago
An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the
Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m

(a) Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2

(b) Magnification, m=\dfrac{h'}{h}

h' is image height

-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m

Hence, this is the required solution.

4 0
3 years ago
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