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3 years ago

What defines a mixture? for sience

1 answer:

8 0

In chemistry, a mixture is a material made up of two or more substances which are not chemically combined.

Hope this helps :)

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Two forces of magnitude 8N and 4N act at right angle to each other. The angle between the resultant and the 8N force is?

Marta_Voda [28]

Answer:

Draw the vector triangle (head to tail)

Let 8 be adjacent and 4 the opposite side

tan theta = 4 / 8 = .5

theta = 26.6 deg

4 0

2 years ago

A ramp is 1.0 m high and 3.0 m long. What is the IMA of the ramp?

oksano4ka [1.4K]

To calculate the ideal mechanical advantage for an inclined plane, divide th length of the incline by the height of the incline.
Therefore; IMA = L/h
L= 3.0 m, while h =1.0 m
IMA = 3/1
= 3
Therefore the IMA of the ramp is 3
This means the ramp increases the force that is being exerted by 3 times.

4 0

3 years ago

There is a refrigerator running in a room, the heat flowing into the refrigerator from the outside is 40 J/s, and the refrigerat

Mamont248 [21]

Answer:

(A) 140 j/sec (b) 1.26 K

Explanation:

We have given the heat heat flowing into the refrigerator = 40 J/sec

Work done = 40 W

(a) So the heat discharged from the refrigerator $=heat\ flowing\ in\ refrigerator+work\ done=40+100=140j/sec$

(b) Total heat absorbed =140 j/sec $=140\times 3600=504000j/hour$

Let the temperature be $\Delta T$

Heat absorbed per hour =504000 $[tex]=400\times 10^3\times \Delta T$

So $\Delta T=\frac{504000}{400000}=1.26K$

8 0

3 years ago

A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

$U=mgh$

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J$

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

$U=mgh$

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

$U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J$

3. 0 J

As before, the gravitational potential energy of the ball is given by

$U=mgh$

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J$

4 0

3 years ago

An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.72 kHz. At what frequency doe

nlexa [21]

Answer:

f'=5.58kHz

Explanation:

This is an example of the Doppler effect, the formula is:

$f'=\frac{(v+v_o)}{(v+v_s)}f$

Where f is the actual frequency, $f'$ is the observed frequency, $v$ is the velocity of the sound waves, $v_o$ the velocity of the observer (which is negative if the observer is moving away from the source) and $v_s$ the velocity of the source (which is negative if is moving towards the observer). For this problem:

$f=3.72kHz\\v=342m/s\\v_o=0m/s\\v_s=-114m/s$

$f'=\frac{(342+0)}{(342-114)}3.72\times10^3\\f'=\frac{342}{228}3.72\times10^3\\f'=(1.5)3.72\times10^3\\f'=5580Hz=5.58kHz$

5 0

3 years ago