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Effectus [21]
2 years ago
13

On the image at right, the two magnets are the same. Which paper clip would be harder to remove?

Physics
1 answer:
serious [3.7K]2 years ago
7 0

Answer:B

Explanation: The book is thinner making magnets attraction stronger, making the paper clip harder to move

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a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,
Effectus [21]
Let h = the distance from the edge of the wall to the water surface (m).

Use g = 9.8 m/s² and neglect air resistance.

The initial vertical velocity of the pebble is zero.
Because the pebble hits the surface of the water after 1.5 s, therefore
h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m

Answer:  11.025 m
7 0
2 years ago
Read 2 more answers
Wind blows at the speed of 30m/s across a 175m^2 flat roof if a house.
Makovka662 [10]

Answer:

the net force is 101587.5 N

Explanation:

The speed of wind

v = 30 m/s

The area of roof,

A = 175 m 2

The expression for the Bernoulli's theorem.

P = 12 ρv 2 ...... (1)

Here,

P is the pressure difference,

ρ is the density of air and

v is the speed of wind.

The expression for the pressure.

P = F A ..... (2)

Here,

F is the force and

A is the area of roof.

Part (a)

Substitute the values for the pressure difference in equation (1)

P = 12 × 1.29 × (30) 2 P = 580.5 Pa

Thus, the pressure difference at the roof between the inside and outside air is

580.5 Pa

Part (b)

Substitute the values for the net force in equation (2)

580.5 = F 175 F = 101587.5 N

Thus, the net force is 101587.5 N.

8 0
2 years ago
The end of a stopped pipe is to be cut off so that the pipe will be open. If the stopped pipe has a total length L, what fractio
Alexxandr [17]

Answer:

4/10 of L.

Explanation:

A stopped pipe is a pipe with one closed end and one opened end. it is also called a closed pipe.

The fundamental mode of a stopped pipe is also called its fundamental frequency, and is f₁=v/4L.

Where f₁=fundamental frequency, v= velocity of sound, L= Length of pipe.

The fifth harmonic of the stopped pipe f₅ =5v/4L .................(1)

For an open pipe,

Fundamental  mode is also called fundamental frequency f₁₀=v/2l₀ .......... (2)

Where f₁₀ = fundamental frequency of a closed pipe, v= velocity of sound and l₀=length of the resulting open pipe.

from the question, the fundamental mode of the resulting open pipe = The fifth harmonic of the original stopped pipe.

∴ f₅=f₁₀

⇒5v/4L = v/2l₀

Equating v from both side of the equation,

⇒ 5/4L = 1/2l₀

Cross multiplying the equation,

5×2l₀ = 4L× 1

10l₀ = 4L

Dividing both side of the equation by the coefficient of l₀ i.e 10

10l₀/10 = 4L/10

∴ l₀ = 4/10(L)

∴ 4/10 of L must be cut off

7 0
2 years ago
In the swing carousel amusement park ride, riders sit in chairs that are attached by a chain to a large rotating drum as shown i
irinina [24]

Answer:\theta =44.068^{\circ}

Explanation:

Given

time taken to complete the circle=7.9 s

radius of circle(r)=15 m

velocity of rider is given by =\frac{2\pi r}{t}

v=\frac{2\pi 15}{7.9}=11.93 m/s

Let us suppose T is the tension in the chain and \thetais the angle which chain makes with vertical

Therefore T\sin \theta =\frac{mv^2}{r}-1

T\cos \theta=mg --2

Divide 1 & 2 we get

tan\theta =\frac{v^2}{rg}

tan\theta =0.968

\theta =44.068^{\circ}

8 0
2 years ago
If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

8 0
2 years ago
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