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2 years ago

Which surface produces the least friction when someone walks on it

Physics

1 answer:

Alika [10]2 years ago

6 0

Answer: B Ice

The friction depends on the texture of the surface. A rough surface has more friction. The carpet or wooden surface that we usually walk on have fairly enough friction. We don't slip on sand also much but to walk on ice it requires spokes as you also see while doing ice skating it is very difficult to stand on ice. This is because of low friction on ice floor.

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Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.

kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees

ELECTRIC FORCE (F)

F = $\frac{KQq}{d^{2} }$

Where K = 9 x $10^{9}$ N $m^{2}$ / $C^{2}$

The distance between $q_{1}$ and $q_{3}$ can be calculated by using Pythagoras theorem.

d = $\sqrt{33^{2} + 33^{2} }$

d = 46.7 cm = 0.467 m

For force $F_{1}$ , substitute all the parameters into the formula above

$F_{1}$ = (9 x $10^{9}$ x 3 x 1)/ $0.467^{2}$

$F_{1}$ = 2.7 x $10^{10}$ /0.218

$F_{1}$ = 1.24 x $10^{11}$ N

For force $F_{4}$ , substitute all the parameters into the formula above

$F_{4}$ = (9 x $10^{9}$ x 3 x 4)/ $0.33^{2}$

$F_{4}$ = 1.08 x $10^{11}$ /0.1089

$F_{4}$ = 9.92 x $10^{11}$ N

For force $F_{2}$ , substitute all the parameters into the formula above

$F_{2}$ = (9 x $10^{9}$ x 3 x 2)/ $0.33^{2}$

$F_{2}$ = 5.4 x $10^{10}$ /0.1089

$F_{2}$ = 4.96 x $10^{11}$ N

Summation of forces on Y component will be

$F_{y}$ = $F_{4}$ - $F_{1}$ Sin 45

$F_{y}$ = 9.92 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{y}$ = 9.04 x $10^{11}$ N

Summation of forces on X component will be

$F_{x}$ = $F_{2}$ - $F_{1}$ Cos 45

$F_{x}$ = 4.96 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{x}$ = 4.08 x $10^{11}$ N

Net Force = $\sqrt{F_{x} ^{2} + F_{y} ^{2} } }$

Net force = $\sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2} }$

Net force = 9.9 x $10^{11}$ N

The direction will be

Tan ∅ = $F_{y}$ / $F_{x}$

Tan ∅ = 9.04 x $10^{11}$ / 4.08 x $10^{11}$

Tan ∅ = 2.216

∅ = $Tan^{-1}$ (2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

8 0

2 years ago

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A car going 22 m/s increases its speed to pass a truck. Five seconds later the car is going 35 m/s. Calculate the acceleration o

jonny [76]

Answer:

I don't know

Explanation:

4 0

2 years ago

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Light in vacuum travels at a speed of 3.00 x 10^8 m ^-1 s^-1 on average earth is 93,000,000 miles from the sun how many minutes

NemiM [27]

The correct unit for the speed of light is [ m s⁻¹ ].

Time = (distance) / (speed)

Time = (9.3 x 10^7 miles) x (1609 m/mile) / (3 x 10^8 m/s) = 498.8 seconds .

That would be <em>8.31 minutes</em>.

6 0

2 years ago

An apparatus like the one Cavendish used to ﬁnd G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.

Ber [7]

Answer:

The magnitude of gravitational force between two masses is $4.91\times 10^{-9}\ N$ .

Explanation:

Given that,

Mass of first lead ball, $m_1=5.2\ kg$

Mass of the other lead ball, $m_2=0.046\ kg$

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

$F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N$

So, the magnitude of gravitational force between two masses is $4.91\times 10^{-9}\ N$ . Hence, this is the required solution.

5 0

2 years ago

Please need in hurry

gulaghasi [49]

Explanation:

i) center of gravity (or mass)

ii) m = W/g = (160 N)/(9.8 m/s^2)

= 16.3 kg

3 0

2 years ago