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4 years ago

Large bodies of water help to keep nearby climates from being too hot or too cold because they

1 answer:

7 0

Large bodies of water help to keep nearby climates from being too hot or too cold because they absorb heat in the summer and slowly release that heat in the winter.

So your answer is absorb

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In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a

SOVA2 [1]

Answer: $0.084 m/s^2$

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance $d_1=8.13 km$

time taken=25 min =1500 s

initial speed $v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s$

after 8.13 km mark steve started to accelerate

speed after 60 s

$v_2=v_1+at$

$v_2=5.6+a\times 60$

distance traveled in 60 sec

$d_2=v_1\times 60+\frac{a60^2}{2}$

$d_2=336+1800 a$

time taken in last part of journey

$t_3=1663.6-1560=103.6 s$

distance traveled in this time

$d_3=v_2\times t_3$

$d_3=\left ( 5.6+a\times 60\right )103.6$

and total distance $=d_1+d_2+d_3$

$10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$a=0.084 m/s^2$

5 0

4 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

Oksi-84 [34.3K]

Answer:

the equation of motion is

$x(t) =-\frac{1}{6} \cos4\sqrt{6} t$

Explanation:

Given that,

The weight attached to the spring is 24pounds

Acceleration due to gravity is 32ft/s²

Assume x is the string length, 4inches

convert the length inches to to feet = 4/12 = 1/3feet

From Hookes law , we calculate the spring constant k

k = W / x

k = 24 / (1/3)

k = 24 / 0.33

k = 72lb/ft

If the mass is displace from its equilibrium position by amount x

the differential equation is

$m\frac{d^2x}{dt^2} + kx=0\\\\\frac{3}{4 } \frac{d^2x}{dt^2}+72x=0\\$

$\frac{d^2x}{dt^2} +96x=0$

Auxiliary equation is

$m^2+96=0\\m=\sqrt{-96} \\m=\_ ^+4\sqrt{6}$

Thus the solution is,

$x(t) = c_1cos4\sqrt{6t} +c_2sin\sqrt{6t}$

$x'(t) =-4\sqrt{6c_1} sin4\sqrt{6t} +c_24\sqrt{6} cos4\sqrt{6t}$

The mass is release from rest

x'(0) = 0

$-4\sqrt{6c_1 } \sin4\sqrt{6} (0)+c_24\sqrt{6} \cos4\sqrt{6} (0)=0\\c_24\sqrt{6} =0\\\\c_2=0$

Therefore

x(t) = c₁ cos4 √6t

x(0) = -2inches

c₁ cos4 √6(0) = 2/12feet

c₁= 1/6feet

There fore, the equation of motion is

$x(t) =-\frac{1}{6} \cos4\sqrt{6} t$

7 0

4 years ago

Why is brainly so dumb? I am currently trying to get into my account right now, and even though I JUST changed it, it keeps sayi

MrRa [10]

Answer:

That happens to me a few times.

Are you sure you are on Brainly.com and not Brainly.Ly?

8 0

3 years ago

Read 2 more answers

A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m

madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂ = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0

4 years ago

Why are bulb filaments made of metals and not wood

saul85 [17]

It's because lightbulbs need to carry an electric current within a range of resistance so that the material doesn't without breaking down.

7 0

3 years ago