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Jobisdone [24]
2 years ago
5

Consider massive gliders that slide friction-free along a horizontal air track. Glider A has a mass of 1 kg, a speed of 1 m/s, a

nd collides with Glider B that has a mass of 5 kg and is at rest. If they stick upon collision, their speed after collision will be ______.
Physics
1 answer:
mamaluj [8]2 years ago
6 0

Answer:

0.167m/s

Explanation:

According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.

Given momentum = Maas × velocity.

Momentum of glider A = 1kg×1m/s

Momentum of glider = 1kgm/s

Momentum of glider B = 5kg × 0m/s

The initial velocity of glider B is zero since it is at rest.

Momentum of glider B = 0kgm/s

Momentum of the bodies after collision = (mA+mB)v where;

mA and mB are the masses of the gliders

v is their common velocity after collision.

Momentum = (1+5)v

Momentum after collision = 6v

According to the law of conservation of momentum;

1kgm/s + 0kgm/s = 6v

1 =6v

V =1/6m/s

Their speed after collision will be 0.167m/s

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The maximum gauge pressure in a hydraulic lift is 18.0 atm what is the largest size vehicle
victus00 [196]
There is a missing part in the question. Found the complete text on internet:
"<span>What is the largest size vehicle (kg) it can lift if the diameter of the output line is 28.0 cm? "

Solution
The diameter of the piston is 28.0 cm, this means its radius is 14.0 cm (half the diameter), so the area of the piston is
</span>A=\pi r^2 = \pi (0.14 m)^2 =6.15 \cdot 10^{-2} m^2
<span>
The maximum pressure of the lift is
</span>p=18.0 atm = 1.82 \cdot 10^6 Pa
<span>
Therefore the maximum force the piston can lift is
</span>F=pA=(1.82 \cdot 10^6 Pa)(6.15 \cdot 10^{-2} m^2)=1.12 \cdot 10^5 N
<span>
And the size (the mass) of the vehicle is
</span>m= \frac{F}{g}= \frac{1.12 \cdot 10^5 Pa}{9.81 m/s^2} =1.14 \cdot 10^4 kg<span>
</span>
3 0
3 years ago
A long copper rod of diameter 2.0 cm is initially at a uniform temperature of 100°C. It is now exposed to an air stream at 20°C
Lisa [10]

Answer:

t = 4.0 min

Explanation:

given data:

diameter of rod = 2 cm

T_1 = 100 degree celcius

Air stream temperature =  20 degree celcius

heat transfer coefficient = 200 W/m2. K

WE KNOW THAT

copper thermal conductivity = k = 401 W/m °C

copper specific heat Cp = 385 J/kg.°C

density of copper = 8933 kg/m3

charateristic length is given as Lc

Lc = \frac{V}{A_s}

Lc = \frac{\frac{\pi D^2}{4} L}{\pi DL}

Lc = \frac{D}{4}

Lc = \frac{0.02}{6} = 0.005 m

Biot number is given as Bi = \frac{hLc}{k}

Bi = \frac{200*0.005}{401}

Bi = 0.0025

As Bi is greater than 0.1 therefore lumped system analysis is applicable

so we have

\frac{T(t) - T_∞}{Ti - T_∞} = e^{-bt} ............1

where b is given as

b = \frac{ hA}{\rho Cp V}

b = \frac{ h}{\rho Cp Lc}

b = \frac{200}{8933*385*0.005}

b = 0.01163 s^{-1}

putting value in equation 1

\frac{25-20}{100-20} = e^{-0.01163t}

solving for t we get

t = 4.0 min

6 0
3 years ago
Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth’s center. Satellite A is to o
Vera_Pavlovna [14]

Answer:

Explanation:

Orbital radius of satellite A , Ra = 6370 + 6370 = 12740 km

Orbital radius of satellite B , Rb = 6370 + 19110 = 25480 km

Orbital potential energy of a satellite = - GMm / r where G is gravitational constant , M is mass of the earth and m is mass of the satellite

Orbital potential energy of a satellite A = - GMm / Ra

Orbital potential energy of a satellite B = - GMm / Rb

PE of satellite B /PE of satellite A

=  Ra / Rb

= 12740 / 25480

= 1 / 2

b ) Kinetic energy of a satellite is half the potential energy with positive value , so ratio of their kinetic energy will also be same

KE of satellite B /KE of satellite A

= 1 / 2

c ) Total energy will be as follows

Total energy = - PE + KE

- P E + PE/2

= - PE /2

Total energy of satellite B / Total energy of A

= 1 / 2

Satellite B will have greater total energy because its negative value is less.

5 0
2 years ago
How many types of electrical charge are there in all materials and what are the charges
JulsSmile [24]

There are two types of electric charges; positive and negative

- If you need more info than this let me know

- hope this helps

4 0
3 years ago
The length of the mercury thread is found to be 4cm and 24cm at ice point and steam point respectively on an ungraduated thermom
BabaBlast [244]

Answer:

The difference between ice and steam in Celsius (Centigrade) is 100 deg.

So the difference between and 4 cm and 24 cm of the thread corresponds to 100 deg C.

So 8 cm is 4 cm greater than the ice point

4 cm / 20 cm = 1/5     since the steam point and the ice point are 20 cm apart

Then 1/5 * 100  deg C = 20 deg C   the requested temperature

6 0
3 years ago
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