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arlik [135]
3 years ago
7

Are soaps salts true or false

Physics
1 answer:
murzikaleks [220]3 years ago
3 0

true

Soaps are water-soluble sodium or potassium salts of fatty acids. Soaps are made from fats and oils, or their fatty acids, by treating them chemically with a strong alkali.

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How much energy must be added to a bowl of 125 popcorn kernels in order for them to reach a popping temperature of 175°C? Assume
Lisa [10]

Answer:

c) 3176 J

Explanation:

mass of 125 popcorn = 125 x .0001 = . 0125 kg

increase in temperature =  175 - 21 = 154°C

specific heat of popcorn = 1650

heat required to increase temperature

= mass of popcorn x increase in temperature x specific heat

= .0125 x 154 x 1650

=  3176.25 J

8 0
3 years ago
This is the third time I’m asking, please, On a wet road, is a higher coefficient of friction on the tires safer or a lower one
jenyasd209 [6]

Answer:

higher is safer

Explanation:

because it is a wet slippery surface, you would need more friction on the tires, to get more traction in the slippery wet road, if you had low friction you would not move anywhere and or could swerve off somewhere

4 0
1 year ago
What is the first step in the event of a fire in a building?
yuradex [85]
I’m assuming D because it’s the most reasonable
8 0
3 years ago
Read 2 more answers
A=100m/s2<br> Vi=10m/s<br> T=10sec<br> Vf?
IrinaK [193]

Answer:

v=u+at

v= 10+100×10

v = 10+1000

v = 1010 m/s

3 0
3 years ago
Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the
Masja [62]

Answer:

distance of 2nd team from 1st team will be:  58.2

Direction of 2nd team from 1st team will be:  14.90 deg North of east

Explanation:

ASSUME Vector is R and  makes angle A with +x-axis,

therefore component of vector R is

R_x = Rcos A

R_y = Rsin A

From above relation

Assuming base camp as the origin, location of 1st team is

R_1 = 37 km away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km

R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km

location of 2nd team is at

R_2 = 32 km, at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km

R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km

Now position of 2nd team with respect to 1st team will be given by:

R_3 = R_2 - R_1

R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j

Using above values:

R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j

R_3 = 51.49 i + 13.71 j

distance of 2nd team from 1st team will be:

\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)

\left | R_3 \right | = 53.28 km = 58.2 km

Direction of 2nd team from 1st team will be:

Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]

Direction = 14.90 deg North of east

6 0
3 years ago
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