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3 years ago

Is better to run or walk when it is raining?

Physics

1 answer:

lawyer [7]3 years ago

5 0

Answer:

run

Explanation:

run - less wet

walking - more wet

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A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the i

drek231 [11]

Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = $\frac{28}{10}$ × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

t = 0.1m × 16N × sin160.4°

t = 0.2845Nm (rounded to 4 decimal places)

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3 years ago

A bus is moving with an acceleration of 4 m/s^2. If it was initially at rest,

shusha [124]

Probably b) 6 m/s
Not sure

6 0

3 years ago

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A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help.

GrogVix [38]

Answer:

$H = 532 m$

Explanation:

When teacher falls into the cliff then she shout for Help at the same time

so here we know that sound will go down and reflect back up

so here in 3 s distance traveled by the sound

$d = vt$

$d = 340 \times 3$

$d = 1020 m$

now in the same time the distance that teacher will fall down is given as

$d_1 = \frac{1}{2}gt^2$

$d_1 = \frac{1}{2}(9.81)(3^2)$

$d_1 = 44.1 m$

now total distance traveled by teacher and sound in 3 s

$d_{total} = d + d_1$

$d_{total} = 1020 + 44.1$

this total distance must be equal to twice the height of the cliff

$2H = 1064.1 m$

$H = 532 m$

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3 years ago

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Which is the correct order of the academic pathway of a pulmonologist?

saul85 [17]

Answer:

Explanation:

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3 years ago

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A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago