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3 years ago

What is the compare the strength of static,sliding,and rolling friction

1 answer:

7 0

The force needed to overcome sliding friction is more than the force needed to overcome rolling friction or static or even fluid

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Am 1 years old UwU (JK 15)

igomit [66]

Answer:

so am i ._. 15 not 1 ofc lol

Explanation:

7 0

2 years ago

An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is

Lady_Fox [76]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}$

$\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

<h3>☯ <u>By using formula of Lens</u> </h3>

$\\$

$\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}$

$\\$

$\dashrightarrow\:\: \sf{ v = 30 \ cm}$

$\\$

<h3>☯ <u>Now, Finding the magnification </u></h3>

$\\$

$\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}$

$\\$

$\dashrightarrow\:\: \sf{m = -2}$

$\\$

<h3>☯ <u>Hence</u>, $\\$ </h3>

$\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}$

$\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}$

3 0

2 years ago

A (20*20) cm² loop has a resistance of 0.10 Ω. A magnetic field perpendicular to the loop is B = 4t - 2t², where B is in tesla a

Ilya [14]

Answer with Explanation:

We are given that

Area of loop= $(20\times 20) cm^2=400\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Resistance, R= $0.1\Omega$

B= $4t-2t^2$

We know that magnetic flux

$\phi=BA$

Emf , $E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid$

Current, $I=\frac{E}{R}$

Current, $I=\frac{\mid 400\times 10^{-4}(4-4t)\mid }{0.1}=1.6\mid (1-t)\mid$

Substitute t=0 s

Then, I= $1.6\mid (1-0)\mid$ =1.6 A

Substitute t=1 s

Then, I= $1.6\mid (1-1)\mid$ =0

Substitute

t=2 s

Current, I= $1.6\mid(1-2)\mid$ =1.6 A

8 0

2 years ago

Read 2 more answers

What is the amplitude of a wave related to

Vanyuwa [196]

Answer:

intensity because square of the amplitude is proportional to the intensity of the wave

Explanation:

3 0

3 years ago

(Q008) Magnetic anomalies are a. found only on the seafloor. b. measured with an instrument called an inclinometer. c. defined a

likoan [24]

Answer: D. places where the magnetic field strength is either greater or less than the expected strength.

Explanation:

Magnetic anomaly refers to the change in the magnitude of the magnetic field of the Eatth with respect to the location's expected value.

Magnetic anomalies are places where the magnetic field strength is either greater or less than the expected strength. Therefore, the correct option is D.

7 0

2 years ago