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3 years ago

A 1000 Kg car traveling at 10 m/s hits the back of a 5000 Kg truck

Physics

1 answer:

insens350 [35]3 years ago

8 0

Answer:

0 m/s

The car becomes stationary

Explanation:

The law of conservation of linear momentum states that the sum of inital and final momentum should be equal and expressed as

$m_cu_c+m_tu_t=m_cv_c+m_tv_t$

Where m represent the mass, u and v are tge initial and final velocities while subscripts c and t represent car and truck.

Taking forward direction as positive then considering that the truck is originally at rest, we substitute original truck velocity with 0, mass of car and truck with 1000 kg and 5000 kg respectively then final truck velocity as 2 m/s as we take initial car velocity to be 10 m/s

1000*10+(5000*0)=5000*2+1000v

1000v=0

V=0

Therefore, the car finally becomes stationary.

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How would you convert steps and jumps to meters?

kumpel [21]

Explanation:

by inches to centimeter u can convert to meters

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3 years ago

In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.

MA_775_DIABLO [31]

A) Francium-223

In an alpha decay, a nucleus decay emitting an alpha particle, which corresponds to a nucleus of helium: so, it consists of 2 protons and 2 neutrons.

$X \rightarrow X' + \alpha$

This means that in the decay:

- The original nucleus loses 2 protons --> so its atomic number Z decreases by 2 units

- The original nucleus loses 2 nucleons (2 protons and 2 neutrons) --> so its mass number A decreases by 4 units

In this example, the original nucleus is Ac (Actinium), with

Z = 89

A = 227

After the decay, it must be

Z - 2 = 89 - 2 = 87

A - 4 = 227 - 4 = 223

We see from the periodict table, Z=87 corresponds to Francium (Fr), so the final nucleus will be francium-223 (the isotope of francium with 223 nucleons).

B) Polonium-211

In a beta-minus decay, a neutron in the nucleus turns into a proton, emitting a fast-moving electron (the beta particle) and an anti-neutrino.

$n \rightarrow p + e^- + \bar{\nu}$

Therefore, in this process:

- The original nucleus gains 1 protons, so its atomic number Z increases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Bi (bismuth)-211, with

Z = 83

A = 211

So After the decay, it will be

Z + 1 = 83 + 1 = 84

A = 211

So, the nucleus will be Polonium (Z=84), isotope with 211 nucleons.

C) Neon-22

In a beta-plus decay, a proton in the nucleus turns into a neutron, emitting a fast-moving positron (the beta particle) and a neutrino.

$p \rightarrow n + e^+ +\nu$

Therefore, in this process:

- The original nucleus loses 1 protons, so its atomic number Z decreases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Na (sodium)-22, with

Z = 11

A = 22

So After the decay, it will be

Z - 1 = 11 - 1 = 10

A = 22

So, the nucleus will be Neon (Z=10), isotope with 22 nucleons.

D) Technetium-98

In a gamma decay, an unstable nucleus emits a gamma ray:

$X' \rightarrow X + \gamma$

In this process, only energy is released (in the form of gamma ray), so there is no gain/loss of protons/neutrons in the process. This means that:

- The atomic number Z remains constant

- The mass number A remains constant

In this example, we have a nucleus of Tc (Technetium)-98, with

Z = 43

A = 98

These numbers will not change during the decay: this means that after the decay, we will still have a nucleus of Technetium-98.

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3 years ago

If lightning is caused by electricity in the air during a thunderstorm, what causes thunder?

NikAS [45]

Answer:

Since light travels faster than sound, the lightning arrives before the thunder. But thunder is the sound of lightning shooting through the sky in a thunderstorm.

Explanation:

7 0

3 years ago

Read 2 more answers

A student places blocks on a 100cm long see-saw as shown/

Hunter-Best [27]

Answer:

Part 1)

$\tau_1 = 5 \times (0.50) = 2.5 N m$

Part 2)

$\tau_2 = 14 \times (0.30) = 4.2 N m$

Part 3)

$\tau_3 = 1.4 N m$

Part 4)

Since torque on right side is more so here it will turn and slip over it

Explanation:

As we know that the block A is placed at distance

d = 50 cm from the hinge at 70 cm mark

So torque due to weight of A is given as

$\tau_1 = 5 \times (0.50) = 2.5 N m$

the block B is placed at distance

d = 30 cm from the hinge at 70 cm mark

So torque due to weight of B is given as

$\tau_2 = 14 \times (0.30) = 4.2 N m$

Now torque due to weight of the scale is given as

$\tau_3 = 7(0.20)$

$\tau_3 = 1.4 N m$

now torque on left side of scale is given as

$\tau_{left} = \tau_1 + \tau_3$

$\tau_{left} = 2.5 + 1.4 = 3.9 N m$

Torque on right Side is given as

$\tau_{right} = \tau_2 = 4.2 Nm$

Since torque on right side is more so here it will turn and slip over it

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3 years ago

A bag is dropped from a balloon that is 50m above the ground and rising at 15m/s. calculate

ANTONII [103]

Answer:

See the answers below

Explanation:

We can solve this problem using the principle of energy conservation. That is, the energy is conserved before and after dropping the bag.

For this case we have mechanical energy, which is the sum of the kinetic and potential energies.

$E_{1}=E_{2}$

where:

$E_{1}=E_{k}+E_{p}\\E_{2}=E_{p}$

Ek = kinetic energy [J] (units of Joules)

Ep = potential energy [J]

In the final Energy (2), there is only potential energy. since when the balloon reaches the maximum height its velocity is zero, that is, there is no kinetic energy.

$m*g*h+\frac{1}{2}*m*v^{2} =m*g*h_{1} \\9.81*50+0.5*(15)^{2}=9.81*h_{1}\\h_{1} = 61.46 [m]$

With the value calculated above we can find the acceleration of the balloon.

The distance traveled is the difference between the maximum height and 50 meters.

$x = 61.46-50\\x = 11.46[m]$

With the following equation of kinematics.

$v_{f}^{2} =v_{o}^{2}+2*a*x$

$0 = 15^{2} +2*a*11.46\\a = - 9.816 [m/s^{2} ]$

The negative sign indicates that the acceleration acts downward. That is, in the opposite direction to the movement.

We can use the following equation of kinematics to find the final velocity after 4 seconds.

$v_{f}=v_{o}-a*t\\v_{f}=15-9.816*(4)\\v_{f}=-24.24 [m/s]$

Now the distance:

$v_{f}^{2} =v_{o}^{2}-2*a*x\\(24.24)^{2} =(15)^{2} -2*9.81*x\\x = 18.48 [m]\\x_{f}=50+18.48 = 68.48 [m]$

c) Using the following equation of kinematics.

$v_{f}=v_{o}-a*t\\0 = 15-9.81*t\\15=9.81*t\\t = 1.52 [s]$

4 0

3 years ago