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3 years ago

7

A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pon

d?

1 answer:

lukranit [14]3 years ago

4 0

A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pond?
It increases!

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Elza [17]

1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum

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3 years ago

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

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3 years ago

220V is applied to two different conductors made of the same material. One conductor is twice as long and twice as thick as the

ollegr [7]

Answer:

Explanation:

For calculating resistance of a conductor , the formula is

R = ρ l / A , ρ is specific resistance , l is length and A is cross sectional area of wire.

For first wire length is l₁ , area is A₁ resistance is R₁, for second resistance is R₂ , length is l₂ and area is A₂

Given , l₁ = 2l₂ , A₁ = 4A₂ , area is proportional to square of thickness.

R₁ / R₂ = I₁A₂ / I₂A₁

= 2l₂ x A₁ / 4 I₂A₁

= 1 / 2

2R₁ = R₂

Power = V² / R

Ratio of power = (V² / R₁) x (R₂ / V²)

= R₂ / R₁

= 2 .

7 0

3 years ago

Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1

Artemon [7]

Answer:

$\frac{F_1}{F_2} =\frac{1}{2}$

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

$F_1=2k\times d$

case 2:

$F_2=2k\times 2d$

therefore, $\frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}$

$\frac{F_1}{F_2} =\frac{1}{2}$

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3 years ago

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A. it provides support

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2 years ago