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4 years ago

7. True or False? The rules governing youth workers is the same for all

Physics

1 answer:

nikitadnepr [17]4 years ago

6 0

True the rules governing youth workers is the same for all youth on the farm or not

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Glycerin at 30°C has a density of 1,260 kg/m3 and a viscosity of 0.630 Pa s. In a laboratory experiment, some glycerin is forced

Alekssandra [29.7K]

Answer:

Explanation:

Rate of flow of liquid through a tube can be expressed by the following expression

V = π P r⁴ / 8ηl

P is pressure difference between end of tube = 618 Pa

r , radius of tube = .5 x 10⁻²

η is viscosity of liquid flowing = .63

l is length of tube = .10 m

V = 3.14 x 618 x ( .5 x 10⁻² )⁴ / (8 x .63 x .10 )

= 240.64 x 10⁻⁸ m³ /s

mass = 240.64 x 1260 x 10⁻⁸ kg / s

= 3.03 x 10⁻³ kg /s

= 3.03 gram /s .

5 0

3 years ago

3. In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 100 m/s collides with a 1.50-kg cart (Cart B) movin

Alex787 [66]

Answer:

The speed of the two carts after the collision is 10 m/s.

Explanation:

Hi there!

The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:

initial momentum = mA · vA1 + mB · vB1

final momentum = (mA + mB) · vAB2

Where:

mA = mass of cart A = 0.500 kg

vA1 = velocity of cart A before the collision = 100 m/s

mB = mass of cart B = 1.50 kg.

vB1 = velocity of cart B before the collision = - 20 m/s

vAB2 = velocity of the carts that move as a single object = unknown.

(notice that we have considered leftward as negative direction)

Since the momentum of system remains constant:

initial momentum = final momentum

mA · vA1 + mB · vB1 = (mA + mB) · vAB2

Solving for vAB2:

(mA · vA1 + mB · vB1) / (mA + mB) = vAB2

(0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2

vAB2 = 10 m/s

The speed of the two carts after the collision is 10 m/s.

6 0

3 years ago

Use the graph above to determine the change in speed of the object between 20 and 30 seconds?

allsm [11]

Answer:

6 m/s

Explanation:

To determine the change in speed of the object, we just need to determine its speed at t = 30 s and at t = 20 s, and then calculate the difference.

The speed at t = 30 is:

v = 6 m/s

While the speed at t = 20 s is:

u = 0

Therefore, the change in speed is:

$\Delta v = v-u=6-0 = 6 m/s$

4 0

3 years ago

Read 2 more answers

A tight knot can be easily opens by using a longer spanner. Give reasons

kobusy [5.1K]

It is because the effort distance is greater than the load distance

Explanation:

As we know, Effort×effort distance = load × load distance

So when effort distance is increases,

The effort decreases

So when the spanner’s handle is long

A tight knot can easily be opened by less effrot

I hope it helped

6 0

2 years ago

Marc attaches a falling 500-kg object with a rope through a pulley to a paddle wheel shaft. He places the system in a well-insul

Klio2033 [76]

Answer:

4.68227 °C

Explanation:

$m_o$ = Mass of object = 500 kg

$m_w$ = Mass of water = 25 kg

c = Specific heat of water at 20°C = 4186 J/kg°C

h = Height from which the object falls = 100 m

g = Acceleration due to gravity = 9.8 m/s²

The potential energy and heat will balance each other

$PE=Q\\\Rightarrowmc m_ogh=m_oc\Delta T\\\Rightarrow \Delta T=\frac{m_ogh}{m_oc}\\\Rightarrow \Delta T=\frac{500\times 9.8\times 100}{25\times 4186}\\\Rightarrow \Delta=4.68227\ ^{\circ}C$

The temperature change in the water is 4.68227 °C

5 0

4 years ago