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3 years ago

High blood levels of LDL cholesterol reduces your risk of developing heart

Physics

2 answers:

SVEN [57.7K]3 years ago

3 0

Answer:

false

Explanation:

stepan [7]3 years ago

3 0

Answer:

False

Explanation:

it does not reduce your risk of developing heart disease 

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Inside a freely falling runaway elevator, your

nasty-shy [4]

The answer is apparent weight is zero.

You are still accelerating downwards at 9.8m/s^2 (if you are on Earth).

You still are being affected by the Earth's gravity.

Not all because of the previous two statements.

Not none because apparent weight is zero as you are falling.

7 0

3 years ago

Read 2 more answers

You throw a 50.0g blob of clay directly at the wall with an initial velocity of -5.00 m/s i. The clay sticks to the wall, and th

Whitepunk [10]

Answer:0.25 kg-m/s

Explanation:

Given

mass of blob $m=50 gm$

initial velocity $u=-5 m/s\ \hat{i}$

time of collision $t=20 ms$

we know Impulse is equal to change in momentum

initial momentum $P_i=mu$

$P_i=50\times 10^{-3}\times (-5)=-0.25 kg-m/s$

Final momentum $P_f=50\times 10^{-3}v$

$P_f=0$ as final velocity is zero

Impulse $J=P_f-P_i$

$J=0-(-0.25)$

$J=0.25 kg-m/s$

5 0

3 years ago

Which wave has a disturbance that is parallel to the wave motion?

Leto [7]

The answer is D.longitudinal

6 0

3 years ago

Moving a neutral wire in a(n) ____ field will induce a(n) _____.

frutty [35]

the answer is b) magnetic current

7 0

3 years ago

An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10

s2008m [1.1K]

Answer:

1.228 x $10^{-6}$ mm

Explanation:

diameter of aluminium bar D = 40 mm

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x $10^{3}$ N

modulus of elasticity E = 85 GN/m^2 = 85 x $10^{9}$ Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = $\frac{\pi D^{2} }{4}$ = $\frac{3.142* 40^{2} }{4}$ = 1256.8 mm^2

area of hole a = $\frac{\pi(D^{2} - d^{2}) }{4}$ = $\frac{3.142*(40^{2} - 30^{2})}{4}$ = 549.85 mm^2

Total contraction of the bar = $\frac{F*L}{AE} + \frac{Fl}{aE}$

total contraction = $\frac{F}{E} * (\frac{L}{A} +\frac{l}{a})$

==> $\frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85})$ = 1.228 x $10^{-6}$ mm

6 0

3 years ago