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3 years ago

9

Hydrogen bonding is a strong attractive force. T OR F

1 answer:

vodomira [7]3 years ago

6 0

Answer:

F

Hope it helps you!

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The value of acceleration due to gravity (g) on a point. 10,000 kilometers above sea level is about. 1.49 meters/second2. How mu

crimeas [40]

We are given with 98 Newton weight of an object on the surface of the earth with an acceleration equal to 9.8 m/s2. This means the mass of the object is equal to 98/9.8 or 10 kg. Hence the weight of the object 10,000 kilometers above sea level where acceleration is 1.49 m/s2 is 14.9 Newtons.

8 0

3 years ago

Read 2 more answers

NEED HELP ASAP!!! Please and thank you

svetlana [45]

Answer:

Always.

Unbalanced forces mean there will be an acceleration on the object, so speed will change no matter how fast its going already.

4 0

3 years ago

A frictionless spring with a 4-kg mass can be held stretched 0.2 meters beyond its natural length by a force of 30 newtons. If t

maxonik [38]

Answer:

The position for mass is $x(t)=36.75sin(37.5t)$

Explanation:

Let x(t) donate the position of mass at time t,Then x satisfies the differential equation

$m\frac{d^2x}{dt^2} +kx=0\\here\\m=4kg\\k=30N/0.2m\\thus\\w^2=k/m=30N/0.2m*4kg=37.5$

The general solution is

$(x)t=C_{1}cos(37 .5t)+C_{2}sin(37.5t)$

It follows

$x'(t)=-37.5C_{1}sin(37.5t)+37.5C_{2}cos(37.5t)\\now\\x(0)=0\\gives\\C_{1}=0\\ and\\x'(0)=1m/s\\gives\\C_{2} =36.75$

thus the position of mass is

$x(t)=36.75sin(37.5t)$

7 0

3 years ago

What is the critical angle of a light beam passing drim a medium(n=2) to a medium (n=1.5)

DIA [1.3K]

Answer:

it is a sad day for rain to fall. Ill answer your question later

Explanation:

6 0

3 years ago

A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 49° w

dsp73

Answer:

a) n2=(n1sin1)(sin2)

b) 1.18

c) 201081632.7m/s

d) 254237288.1m/s

Explanation:

a) We can calculate the index of refraction of the second material by using the Snell's law:

$n_1sin\theta_1=n_2sin\theta_2$

$n_2=\frac{n_1sin\theta_1}{sin\theta_2}$

b) By replacing in the equation of a) we obtain:

$n_2=\frac{(1,47)sin49\°}{sin69.5\°}=1.18$

c) light velocity in the medium is given by:

$v=\frac{c}{n_1}=\frac{3*10^{8}m/s}{1.47}=204081632.7\frac{m}{s}$

d)

$v=\frac{c}{n_2}=\frac{3*10^{8}m/s}{1.18}=254237288.1\frac{m}{s}$

hope this helps!!

4 0

3 years ago