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3 years ago

A hydraulic system is used to lift a 20,000 N vehicle in an auto garage. If the vehicle sits on

Physics

1 answer:

ollegr [7]3 years ago

7 0

Answer:

Explanation:

Area covered by the vehicle on the piston A1 = 0.5 m2 Weight of the vehicle, F1 = 2000 kg × 9.8 m s-2 Area on which force F2 is applied, A = 0.03 m2 P1 = P2 ; F 1 A 1 F1A1 = F 2 A 2 F2A2 and F2 = F 1 A 1 F1A1 A2 ; F2 = (2000 × 9.8) 0.03 0.5 0.030.5 = 1176 NRead more on Sarthaks.com - https://www.sarthaks.com/957542/hydraulic-system-used-lift-2000-vehicle-in-an-auto-garage-if-the-vehicle-sits-on-piston-area

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6th grade science I mark as brainliest.

DerKrebs [107]

Answer:

2m 13 $\frac{1}{3}$ s

Explanation:

1.5m = 1s

200m = $\frac{200}{1.5}$ × 1s

= 133 $\frac{1}{3}$ s

= 2m 13 $\frac{1}{3}$ s

3 0

2 years ago

Three small balls of the same size but different masses are hung side-by-side in parallel on the strings of same length. They to

andrey2020 [161]

Answer:

m1/6 ( c )

Explanation:

since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy

After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3 ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3 ----- ( 3 )

All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3

note ; 3p = m1v ∴ m3 = $\frac{m1v}{3v3}$ ----- ( 4 )

applying conservation of energy

3v = v1 + v2 + v3 ------- ( 5 )

also 3m1v1 = m1v = v1 = v/3 =

v2 + v3 = 8/3 v ----- ( 6 )

next eliminate V3 for equation 6 by applying conservation of energy and momentum

m1 = 2m2 ------ ( 7 )

now using p1 = p2 = m1v1 = 1/2 m1v1 hence v2 = 2v1 where v1 = 1/3 v

hence ; v2 = 2/3 v ------- ( 8 )

solving with equation 6 and 8

v3 = 2v ------ ( 9 ) ∴ v/v3 = 1/2 ---- ( 10 )

solving with equation 9 and 10

m3 = m1/3 * 1/2 = m1/6

8 0

3 years ago

What happens to the mass and volume if density increases

nikitadnepr [17]

They all stay the same regardless

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3 years ago

How does the radius of a string affect centripetal force.

KiRa [710]

Answer:

because a raduis is half of 25% of a cicrle.

Explanation:

5 0

2 years ago

You lift a 25-kg child 0.80 m, slowly carry him 10 m to the playroom, and finally set him back down 0.80 m onto the playroom flo

maksim [4K]

To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations

$\text{Mass of the child} = m = 25kg$

$\text{Acceleration due to gravity} = g = 9.81m/s^2$

$\text{Height lifted} = h = 0.80m (Upward)$

Work done to upward the object

$W = mgh$

$W = (25)(9.81)(0.8)$

$W = 196.2J$

Horizontal Force applied while carrying 10m,

$F = 0N$

$W = 0J$

Height descended in setting the child down

$h' = -0.8m (Downwoard)$

$W = mgh'$

$W = (25)(9.81)(-0.80)$

$W = -196.2J$

For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.

6 0

3 years ago