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Wewaii [24]
2 years ago
13

А masdOf 500kg a raised to a height of 6m In 30sFind (a) Workdone .​

Physics
1 answer:
Mazyrski [523]2 years ago
5 0

Answer:

Work done is 882000joule.

power is 29400watt.

Explanation:

given,

Mass(m)=500kg

Acceleration due to gravity(g)=9.8m/s²

Height(h)=6m

Time taken(t)=30s

Workdone=?

Power=?

now,

workdone=force*displaxement

= m*g*h

=500*9.8*6

=8,82,000joule

so, the work done by the man is 8,82,000joule.

then,

power=workdone/time taken

=8,82,000/30

=29,400watt

so, the required power to lift a load is 29,400watt.

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An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10
STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi(\frac{6894.76\ Pa}{1\ psi}) = 379212 Pa

\rho = density of water = 1000 kg/m³

Therefore,

v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

\frac{V}{t} = Av\\\\t =\frac{V}{Av}

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = \frac{\pi (0.015875\ m)^2}{4} = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = [\frac{\pi (9.144\ m)^2}{4}][1.524\ m] = 100.1 m³

Therefore,

t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

7 0
2 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Natasha_Volkova [10]

Ok srry bout last time but the answer is A) kelvin: time the reason is because everything else is a SI. Hope this helps and TURTLE

7 0
3 years ago
Read 2 more answers
two test cars of equal mass moving towards each other collide on a horiontal frictionless surface. Before the collision, car A h
LiRa [457]

Answer:

4 m/s

Explanation:

m1 = m2 = m

u1 = 20 m/s, u2 = - 12 m/s

Let the speed of composite body is v after the collision.

Use the conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

m x 20 - m x 12 = (m + m) x v

20 - 12 = 2 v

8 = 2 v

v = 4 m/s

Thus, the speed of teh composite body is 4 m/s.

4 0
2 years ago
Question 1 (1 point)
deff fn [24]

Answer:

Travelled 18 km, they are 6 km from home.

Explanation:

12/2 (halfway) is 6km. So, 6 + 12 would be 18 km, total amount travelled. The total distance of the trip would be 24 km (12 km out, 12km back) if they travelled 12+6 (18km) then they only have 6 km more to go.

5 0
3 years ago
Would it be true that if you double the distance of an astronaut from a planet, the gravitational pull between them would be hal
velikii [3]

Answer:

Yes

Explanation:

Newton's law of universal gravitation is usually stated that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses(m1 and m2) and inversely proportional to the square of the distance between their centers(r).

F = Gm1m2/r²

This is a general physical law derived from

empirical observations by what Isaac Newton called inductive reasoning.

when distance is doubled the gravitational force will be reduced by quarter not half.

5 0
3 years ago
Read 2 more answers
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