15 cm

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

3 years ago

It's not D

DanielleElmas [232]

A. it provides support

5 0

2 years ago

A 5.4 g lead bullet moving at 261 m/s strikes a steel plate and stops. If all its kinetic energy is converted to thermal energy

scZoUnD [109]

Answer:

Change in temperature ∆(tita) is 266.097°C

Explanation:

Ok kinectic energy = 1/2MV²

5.4 grams =( 5.4/1000) kilogram

Kinectic energy =( 1/2 )*(5.4/1000)*261²

Kinectic energy = 183.9267 joules

If kinetic energy = thermal energy

183.9267 joules = mc∆(tita)

Where ∆(tita) = change in temperature

And c = 128 J/kg

∆(tita) = 183.9267/((5.4/1000)*128)

∆(tita) = 266.097

∆(tita) = 266.097°C

8 0

3 years ago

1. A strobe pattern is made of marks on a piece of paper recorded every 0.1 s. Two adjacent marks near the middle of the pattern

Korvikt [17]

Speed = distance / time

3.4cm / 0.1s = 34 cm/sec

7 0

2 years ago

A car with a mass of 1.1 x10 to the third power killograms hits a stationary truck with a mass of 2.3kg x10 to the third power f

Anton [14]

Answer:

v2f = +15.8 m/s

Explanation:

Conservation law of linear momentum:

m1v1i + m2vi2 = m1v1f + m2v2f

Given:

m1 = 1.1 × 10^3 kg

m2 = 2.3 × 10^3 kg

v1i = +22.0 m/s

v2i = 0

v1f = -11.0 m/s

v2f = ?

Re-arranging the conservation law, we get

m1v1i = m1v1f + m2v2f

Solving for v2f,

m2v2f =m1(v1i - v1f)

or

v2f = m1(v1i - v1f)/m2

= (1.1 × 10^3 kg)(22.0 m/s - (-11.0 m/s))/(2.3 m/s)

= (1.1 × 10^3 kg)(33.0 m/s)/(2.3 × 10^3 kg)

= +15.8 m/s

5 0

3 years ago