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3 years ago

When an automotive battery is fully charged, the sulfuric acid and water mixture will have a specific gravity of about

Physics

2 answers:

kari74 [83]3 years ago

8 0

Answer:1.3

Explanation:

Took the test on Penn foster

siniylev [52]3 years ago

6 0

When an automotive battery is fully charged, the mixture of water
and sulfuric acid in it has a specific gravity of about 1.3 .

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A4 kg bowling ball begins rolling down a at bowling alloy at 6 m/s . When it strikes the pins, it is estimated to be moving at 5

Paul [167]

Answer:

Energy lost due to friction is 22 J

Explanation:

Mass of the ball m = 4 kg

Initially velocity of ball v = 6 m/sec

So kinetic energy of the ball $KE=\frac{1}{2}mv^2$

$KE=\frac{1}{2}\times 4\times 6^2=72J$

Now due to friction velocity decreases to 5 m/sec

Kinetic energy become

$KE=\frac{1}{2}\times 4\times 5^2=50J$

Therefore energy lost due to friction = 72 -50 = 22 J

8 0

3 years ago

Hey can anyone please help me with this it’s due in few hours and I’m stuck with ittt

Ray Of Light [21]

Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about $10 ms^-^2$ , the density of whatever you're sinking in, and the depth at which you are. In formula, $p(h) = p_0 + \rho g h$ , and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - $1000 * 10 * 1 = 10^4 Pa$ , while in B is $1000*10*2 = 2*10^4 Pa$ , so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is $800*10*2 = 1,6*10^4 Pa$ : Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so $4m*3m*2m = 24m^3$

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: $800* 24 = 1,92 *10^4kg$

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's $1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N$

2d. $\rho gh$ again, what a surprise! $800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa$

3. Yet again, $\rho gh$ . $1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa$

4 0

2 years ago

If you stood on a planet with four times the mass of Earth, and twice Earth's radius, how much would you weigh?

nikdorinn [45]

Answer:

1/4 times your earth's weight

Explanation:

assuming the Mass of earth = M

Radius of earth = R

∴ the mass of the planet= 4M

the radius of the planet = 4R

gravitational force of earth is given as = $\frac{GM}{R^{2} }$

where G is the gravitational constant

Gravitational force of the planet = $\frac{G4M}{(4R)^{2} }$

= $\frac{G4M}{16R^{2} }$

= $\frac{GM}{4R^{2} }$

recall, gravitational force of earth is given as = $\frac{GM}{R^{2} }$

∴Gravitational force of planet = 1/4 times the gravitational force of the earth

you would weigh 1/4 times your earth's weight

3 0

3 years ago

Jerry runs 60 meters east and then 20 meters west in 10

liraira [26]

Answer: 8 meters per second

Explanation: If you add 60 to 20 you get 80 meters and since he ran those 80 meters in 10 seconds you divide 80 by ten and get 8 and then you get 8m/s

4 0

3 years ago

Question 4 How much time does it take to walk 8 km north at a velocity of 3.8 km/h?

IrinaVladis [17]

Given parameters:

Displacement = 8km

Velocity = 3.8km/h

Unknown:

time = ?

Solution:

Velocity is displacement divided by time.

Velocity = $\frac{displacement}{time}$

Displacement = velocity x time

Input the parameters:

8 = 3.8 x time

Time = $\frac{8}{3.8}$ = 2.1s

The time taken is 2.1s

6 0

3 years ago