All categories

2 years ago

Superman stops a "speeding locomotive" of mass 8.0x103 kg in 4.0 seconds. If the train was originally moving at 40.m/s,

Physics

1 answer:

madreJ [45]2 years ago

8 0

Hello!

a) What is the change in momentum?

The change of the momentum is the velocity, because the velocity is reduced to zero.

We can calculate the aceleration, applicating the formula:

$\boxed{a=\frac{V-Vi}{t} }$

Like you see, the final velocity will be zero, so:

$a = \dfrac{0m/s-40m/s}{4s}$

$a = \dfrac{-40m/s}{4s}$

$a = -10\ m/s^{2}$

Like you see, the aceleration applicate by superman is of -10 m/s^2.

If the aceleration is negattive, means the velocity will decrease.

b.) What is the magnitud of the force wich superman exerts on the train?

For calculate the force applicate for superman, lets applicate the second law of Newton:

$\boxed{F=ma}$

Lets replace and resolve it:

F = 8x10^3 kg * (-10 m/s^2)

F = 8000 kg * (-10 m/s^2)

F = -80 000 N

The force applicated is of -80000 Newtons.

When the force is negative means the force applicated in the another direction of the object what is traveling.

You might be interested in

How do you find speed of a ball at maximum height

Crazy boy [7]

Explanation:

At maximum height, the vertical component of a ball's speed is 0. So the net speed is equal to whatever the horizontal component of the ball is. If the ball is launched straight up, the speed at the highest point is simply 0.

5 0

3 years ago

On a safari, a team of naturalists sets out toward a research station located 4.63 km away in a direction 38.7 ° north of east.

Talja [164]

This is a vector subraction problem.

5 0

3 years ago

Read 2 more answers

Click on the boxes below to indicate the number of electrons or protons in each labeled position for a zinc atom (z=30, a=65). P

g100num [7]

Answer:

Name: Zinc

Symbol: Zn

Atomic Number: 30

Atomic Mass: 65.39 amu

Melting Point: 419.58 °C (692.73 K, 787.24396 °F)

Boiling Point: 907.0 °C (1180.15 K, 1664.6 °F)

Number of Protons/Electrons: 30

Number of Neutrons: 35

Classification: Transition metal

Crystal Structure: Hexagonal

Density at 293 K: 7.133 g/cm3

Color: bluish

('lil long, sorry)

7 0

2 years ago

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

2 years ago

What are three types of electromagnetic waves that are used to transmit information? What types of devices are used to receive e

MrMuchimi

Answer:

Three types of electromagnetic waves, used to transmit various information

Explanation:

A form of energy waves having both electric & magnetic fields are Electromagnetic waves. Three types -

Radio Waves - These have longest wavelengths & transmit data through radio, satellites, radar .

Micro Waves - These have shorter wavelengths & are used in cooking appliances & predicting weather.

X rays - These have more short wavelength and can penerate soft tissues like skin & muscle, hence are used for medical examining

3 0

2 years ago