All categories

3 years ago

5. Emanuel Zacchini, the famous human cannonball of the Ringling Bros. and Barnum and Bailey Circus,

Physics

1 answer:

Strike441 [17]3 years ago

3 0

Explanation:

सिद्ध कीजिए किसी भी बराबर भुजाओं वाले त्रिभुज में उनके सामने के कोण बराबर होते है

You might be interested in

a toy train set has a resistance of 20 ohms and uses a current of 250mA. If it ran for one hour, what is the power of the train?

postnew [5]

Wmwiiejejejejejjeuwuwuwiwwvhwbwvsbe shshe

4 0

3 years ago

Two manned satellites approaching one another at a relative speed of 0.150 m/s intend to dock. The first has a mass of 2.50 ✕ 10

Reika [66]

Answer: $u_{1}=-0.075m/s$ and $u_{2}=0.500m/s$

Explanation:

An elastic collision is one in which both the total kinetic energy of the system and the linear momentum are conserved. That is, during the collision there is no sound, heat or permanent deformations in the bodies as a result of the impact.

Now, in the case of the satellites described here, we have:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1}+m_{2}u_{2}$ (1) Conservation of momentum

$\frac{1}{2}m_{1}v_{1}^{2} +\frac{1}{2}m_{2}v_{2}^{2} =\frac{1}{2}m_{1}u_{1}^{2} +\frac{1}{2}m_{2}u_{2}^{2}$ (2) Conservation of kinetic energy

Where:

$m_{1}=2.5(10)^{3}kg$ is the mass of the first satellite

$m_{2}=7.5(10)^{3}kg$ is the mass of the second satellite

$v_{1}=0.150m/s$ is the initial velocity of the first satellite

$v_{2}=0m/s$ is the initial velocity of the second satellite (we are told it is at rest)

$u_{1}$ is the final relative velocity of the first satellite

$u_{2}$ is the final relative velocity of the second satellite

Now, as we know the second satellite is at rest before the collision, equations (1) and (2) change to:

$m_{1}v_{1}=m_{1}u_{1}+m_{2}u_{2}$ (3)

$\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{1}u_{1}^{2} +\frac{1}{2}m_{2}u_{2}^{2}$ (4)

Solving this system of equations we have the equations for $u_{1}$ and $u_{2}$ :

$u_{1}=\frac{v_{1}(m_{1}-m_{2})}{m_{1}+m_{2}}$ (5)

$u_{2}=\frac{2m_{1}v_{1}}{m_{1}+m_{2}}$ (6)

Substituting the known values on both equations:

$u_{1}=\frac{0.150m/s(2.5(10)^{3}kg-7.5(10)^{3}kg)}{2.5(10)^{3}kg+7.5(10)^{3}kg}$ (7)

$u_{1}=-0.075m/s$ (8) This is the final relative velocity of the first satellite

$u_{2}=\frac{2(2.5(10)^{3}kg)(0.150m/s)}{2.5(10)^{3}kg+7.5(10)^{3}kg}$ (9)

$u_{2}=0.500m/s$ (10) his is the final relative velocity of the second satellite

7 0

4 years ago

A lift travelling up to the top floor of the Empire State building with a mass of 4200kg and a kinetic energy of 4116J. find the

Tomtit [17]

Answer:

$v = 1.4 m/s$

(v= velocity)

Explanation:

Kinetic Energy is computed using the formula

$KE = \dfrac{1}{2}mv^2$

where m = mass of the object in kg and v is the velocity in m/s

Given KE = 4116J,
m = 4200 kg

Substitute and solve for v

$4116 = \dfrac{1}{2}4200\cdot v^2$

=> $4116 = 2100\cdot v^2$

$v^2 = \dfrac{4116}{2100}\\\\v^2 = 1.96\\\\v = \sqrt{1.96}\\\\v = 1.4 m/s$

6 0

2 years ago

Which of the following is true regarding units of geologic time?

Alex Ar [27]

A. is the true answer

4 0

3 years ago

Read 2 more answers

A certain lightbulb has a tungsten filament with a resistance of 20.3 Ω when at 20.0°C and 135 Ω when hot. Assume the resistivit

Paladinen [302]

Answer:

1275.61° C

Explanation:

The resistance of a conductor varies as a function of temperature and it is given as follows

$R= R_o[1+\alpha(T-T_0)]$

R_0 is the resistance at temperature T_0, R is the resistance at T and α is the coefficient of temperature of resistance

α= 0.0045/°C

R_0 = 20.3Ω T_0 = 20°C and R= 135Ω

$135= 20.3[1+0.0045(T-20)]$

T= 1275.61° C

Therefore the temperature of the hot filament = 1275.61° C

3 0

4 years ago