What is responsible for creating distinct weather patterns?

Two identical twins hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about the

777dan777 [17]

Answer:

Part a)

$L = 2683.2 kg m^2/s$

Part b)

$v' = 8.60 m/s$

Part c)

$W = 4326.7 J$

Explanation:

Part a)

As we know that there is no external torque on the system of two twins

so here we will use

$L = mv r + mvr$

$L = 2(78 \times 4.30 \times 4)$

$L = 2683.2 kg m^2/s$

Part b)

Since angular momentum is conserved here as there is no external torque

so we will have

$2(m v r) = 2( m v' \frac{r}{2})$

$v' = 2v$

$v' = 8.60 m/s$

Part c)

Work done by both of them = change in kinetic energy

so we have

$W = 2(\frac{1}{2}mv'^2 - \frac{1}{2}mv^2)$

$W = m(v'^2 - v^2)$

$W = 78(8.60^2 - 4.3^2)$

$W = 4326.7 J$

5 0

3 years ago

A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,

Doss [256]

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a) Kinetic energy of block = potential energy in spring

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg. Spring constant k is unknown, but you can find it from given data:

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.

From the energy equation above, solve for v,

v = v √(k/m)

= 0.15 √(300/1)

= 2.598 m/s.

b) Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.

This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

6 0

3 years ago

In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. The height of th

Tamiku [17]

Answer:

2.124 kg of water

Explanation:

height of the falls is about 48 meters.

Mass of water needed is 1kg = 1000g

Power needed is 106 watts.

The amount of energy in 106 watts in one sec is 106 joules.

To calculate the energy of the 1kg falling water = Mgh

Energy = 1000*9.81*48

Energy = 470880 joules.

1 megawatt is = 1000000watts

The kilogram of water needed is 1000000/470880 = 2.124 kg of water

3 0

3 years ago

A string has a mass of 0.8Kg. The length of the string is 2m. The tension in the string is 5N. The string is stretched and fixed

viktelen [127]

Answer:

1.25Hz

Explanation:

For waves on a string, the second harmonic is obtained from;

2fo = 1/l √T/M

Where;

l = length of the string

M= mass in kilograms

T = tension in the string

2fo = 1/2√5/0.8

2f0 = 1.25Hz

3 0

2 years ago

a 58 kg sprinter, starting from rest, runs a 60 m long race in 7.8 s with a constant acceleration. how much work does the sprint

Alex777 [14]

Work done by the sprinter in first 2.5 s will be 702.699 J.

Work done is the product of force and distance covered.

Mathematically, W = Force * Distance

Mass of the sprinter = 58 kg

Distance covered by the sprinter = s = 60 m

Total time taken by the sprinter = t = 7.8 s

According to the Equations of motion S = ut + $\frac{1}{2}at^{2}$

u = initial velocity = 0

a = constant acceleration

60 = $\frac{1}{2}a7.8^{2}$

60.84 a = 60 * 2

a = 1.97 $m/s^{2}$

For a = 1.97 $m/s^{2}$ and t = 2.5 s

d = $\frac{1}{2}at^{2}$

Here also u is '0' because the distance is being measured from the starting point.

d = $\frac{1}{2} *1.97*(2.5)^{2}$

d = 6.15 m

Work done by the sprinter = F * d

Force = mass * acceleration

F = 58 * 1.97

F = 114.26 N

distance for first 2.5 sec is 6.15 m

Therefore Work done = 6.15 * 114.26

W = 702.699 J

Work done by the sprinter is 702.699 J.

To know more about Work done,

brainly.com/question/28338228

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8 0

1 year ago