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2 years ago

Differences between looping and simersaulting

1 answer:

3 0

Somersaulting- for longer distances.It bends the narrow end in the direction it wants to go & takes grip with tentacles. It releases the broad end and straightens up. like this it continues. looping- for shorter distances.

Hope this helps

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A 55.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300. What h

MariettaO [177]

Answer:

161.86 N

Explanation:

mass of box m= 55.0 kg

weight of the box, mg= 55×9.81

g here is acceleration due to gravity =9.81 m/sec^2

coefficient of friction between the box and the surface μ= 0.3

the friction force F_s= μmg= 0.3×55×9.81

=161.86 N

to move the ball horizontal force required is 161.86 N

8 0

2 years ago

The concentration of an acid or base refers to how completely it dissociates in

Novosadov [1.4K]

Answer:

False

Explanation:

The Concentration of Acid or Base is the ph of the solution.

6 0

2 years ago

Indigenous people sometimes cooked in watertight baskets by placing enough hot rocks into the water to bring it to a boil. What

yaroslaw [1]

Answer:

The rock has a mass of 4.02 kg

Explanation:

Step 1: Data given

Mass of the rock = TO BE DETERMINED

Temperature of the rock = 500 °C

Mass of the water =4.24 kg

⇒ loses 0.044kg as vapor

Initial temperature of the water = 29°C

Final temperature = 100°C

Specific heat of rock = 0.20 kcal/kg °C

Specific heat of water = 1kcal/kg°C

Latent heat of vaporization = 539 kcal/kg

Step 2: formules

Qlost,rock + Qgained,water = 0

Qtotal,water = Qwater +Qvapor

Step 3: Calculate Qvapor

Qvapor = mass of vapor * Latent heat of vapor

Qvapor = 0.044kg * 539 kcal/kg = 23.716 kcal

Step 4: Calculate Qwater

Qwater = mass of water * specific heat * Δtemperature

Qwater = 4.196 kg * 1kcal/kg°C *( 100-29)

Qwater = 297.916 kcal

Step 5: Calculate Qwater,total

Qwater,total = Qwater + Qvapor

Qwater,total = 23.716 kcal + 297.916 = 321.632 kcal

Step 6: Calculate Qrock

Qrock = mass of rock * specific heat rock * Δtemperature

Qrock = mass of rock * 0.20 kcal/kg°C * (100-500)

Qrock = mass of rock * -80 kcal/kg

Step 7: Calculate mass of rock

Qlost,rock + Qgained,water = 0

Qlost,rock = -Qgained,water

mass of rock * -80 kcal/kg = -321.632 kcal

mass of rock = 4.02 kg

The rock has a mass of 4.02 kg

7 0

2 years ago

A vehicle moving with a uniform a acceleration of 2m/s and has a velocity of 4m/s at a certain time .

Evgesh-ka [11]

Answer:

1 second later the vehicle's velocity will be:

$v(1)= 6\,\,\frac{m}{s} \\$

5 seconds later the vehicle's velocity will be:

$v(5)=14\,\,\frac{m}{s}$

Explanation:

Recall the formula for the velocity of an object under constant accelerated motion (with acceleration " $a$ "):

$v(t)=v_0+a\,t$

Therefore, in this case $v_0=4\,\,\frac{m}{s}$ and $a=2\,\,\frac{m}{s^2}$

so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:

$v(t)=v_0+a\,t\\v(t)=4+2\,\,t\\v(1)=4+2\,(1) = 6\,\,\frac{m}{s} \\v(5)=4+2\,(5)=14\,\,\frac{m}{s}$

3 0

2 years ago

Water (2510 g ) is heated until it just begins to boil. if the water absorbs 5.01×105 j of heat in the process, what was the ini

natka813 [3]

E=energy=5.09x10^5J = 509KJ
M=mass=2250g=2.25Kg 
C=specific heat capacity of water= 4.18KJ/Kg 
ΔT= change in temp= ? 
E=mcΔT 
509=(2.25)x(4.18)xΔT 
509=9.405ΔT 
ΔT=509/9.405=54.1degrees 
Initial temp = 100-54 = 46 degrees 
Hope this helps :)

5 0

3 years ago

Differences between looping and simersaulting​

Differences between looping and simersaulting