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FinnZ [79.3K]
3 years ago
11

Kinetic Energy RE-

Physics
1 answer:
tatiyna3 years ago
8 0

Answer:

1) The potential energy, P.E  = 4527.6 J

2) The mass, m = 1.28 kg

3) The height of the bell, h = 4.31 m

4)  The kinetic energy, K.E = 385 J

5) The velocity of the car, V = 22.53 m/s

6) The mass of the car,  m = 17.59 kg

Explanation:

1) The height of the hill, h = 21 m

The carriage with a baby has mass of, m = 22 kg

The potential energy,

                             P.E = mgh

                                   = 22 x 9.8 x 21

                                  = 4527.6 J

2) The height of the platform, h = 20 m

The P.E of the cinder block, P.E = 250 J

The mass,

                                m = P.E / gh

                                     = 250 / (9.8 x 20)

                                    = 1.28 kg

3) The mass of the bell, m = 18 kg

The P.E of the bell, P.E = 760 J

The height of the bell,

                            h = P.E/mg

                               = 760 / 18 x 9.8

                               = 4.31 m

4) The mass of the runner, m = 55 kg

The velocity of the runner, v = 14 m/s

The kinetic energy,

                              K.E = ½ mv²

                                     = ½ x 55 x 14

                                     = 385 J

5)The kinetic energy of the car, K.E = 69,759 J

The mass of the car, m = 275 kg

The velocity of the car,

                                 V = √(2K.E/m)

                                  V = √( 2 x 69759 / 275)

                                      = 22.53 m/s

6) The velocity of the car, v = 38 m/s

The kinetic energy of the car, K.E = 12700 J

The mass of the car,

                                m = 2 K.E/ v²

                                     = 2 x 12700 / 38²      

                                 m = 17.59 kg

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Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured
Mamont248 [21]

Answer:

74.86°C

Explanation:

P₂ = Vapour pressure of water at sea level = 760 mmHg

P₁ = Pressure at base camp = 296 mmHg

T₂ = Temperature of water = 373 K

ΔH°vap for H2O = 40.7 kJ/mol = 40700 J/mol

R = Gas constant = 8.314 J/mol K

From Claussius Clapeyron equation

ln\frac{P_2}{P_1}=\frac{\Delta H}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\\\Rightarrow ln\frac{760}{296}=\frac{40700}{8.314}\left(\frac{1}{T_1}-\frac{1}{373}\right)\\\Rightarrow ln\frac{760}{296}\times \frac{8.314}{40700}+\frac{1}{373}=\frac{1}{T_1}\\\Rightarrow 0.0028735=\frac{1}{T_1}\\\Rightarrow T_1=347.996\ K

T₁ = 347.996 K = 74.86°C

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4 0
3 years ago
A 3.9 g dart is fired into a block of wood with a mass of 24.6 g. The wood block is initially at rest on a 1.5 m tall post. Afte
Galina-37 [17]

Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses m_1 and m_2 moving with velocities u_1 and u_2 before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

m_1u_1+m_2u_2=(m_1+m_2)v..................(1)

were v is their common velocity after impact. If the second mass m_2 was at rest before the impact, then its initial velocity u_2=0m/s. therefore m_2u_2=0. Equation (1) then becomes;

m_1u_1=(m_1+m_2)v..............(2)

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?

Substituting these values into (2), we get the following;

0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

R= vt............(4)

were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;

h=\frac{1}{2}gt^2...........(5)

were g is acceleration due to gravity taken as 9.8m/s^2. Therefore;

1.5=\frac{1}{2}*9.8*t^2\\1.5=4.9t^2\\t^2=\frac{1.5}{4.9}=0.306\\t=\sqrt{0.306} =0.55s

We then substitute R and t into equation (4) to obtain v.

3.5=v*0.55\\v=\frac{3.5}{0.55}\\v=6.36m/s

We now further substitute this value of v into (3) to obtain u_1;

u_1=\frac{0.0285v}{0.0039}\\\\u_1=\frac{0.0285*6.36}{0.0039}\\\\u_1=\frac{0.18126}{0.0039}\\\\u_1=46.48m/s

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