1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
diamong [38]
3 years ago
7

How much energy or stopping power is needed to bring a car to a stop from 100 mph?

Physics
1 answer:
Fed [463]3 years ago
3 0
I think 100 mph pushing the car the opposite direction
You might be interested in
Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th
Kipish [7]

Answer:

Probability of tunneling is 10^{- 1.17\times 10^{32}}

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J

Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

Thus

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = 1.52\times 10^{-35}\ m

Now,

We can calculate the tunneling probability as:

P(t) = e^{\frac{- 2t}{\eta}}

P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}

P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

6 0
3 years ago
What effects are jets and magnetic fields thought to have on a protostar?
alexgriva [62]

Answer:

the effects that a jet and the magnetic fields have on a ProStar Is :

Explanation:

over the years scientist have found out that magnetic fields and the Jets carry away angular momentum , which helps the ProStar grow more in size .

7 0
3 years ago
Atmospheric pressure varies from day to day. The level of a floating ship on a high-pressure day is (a) higher (b) lower, or (c)
frez [133]

Answer:

(c) no different than on a low-pressure day.

Explanation:

The force acting on the ship when it floats in water is the buoyant force. According to the Archimedes' principle: The magnitude of buoyant force acting on the body of the object is equal to the volume displaced by the object.

Thus, Buoyant forces are a volume phenomenon and is determined by the volume of the fluid displaced.  

<u>Whether it is a high pressure day or a low pressure day, the level of the floating ship is unaffected because the increased or decreased pressure at the all the points of the water and the ship and there will be no change in the volume of the water displaced by the ship.</u>

5 0
3 years ago
Tyler's favourite number is 8642, and he tries to fit it into everything he does. If he wishes to do that much work by climbing
Brums [2.3K]

Answer:

2014

Explanation:

none

6 0
3 years ago
The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t
dem82 [27]

Answer:

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

T = 2\pi \sqrt{\frac{r^3}{GM}}

now for the time period of moon around the earth we can say

T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}

here we know that

T_1 = 0.08 year

r_1 = 0.0027 AU

M_e = mass of earth

Now if the same formula is used for revolution of Earth around the sun

T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

here we know that

r_2 = 1 AU

T_2 = 1 year

M_s = mass of Sun

now we have

\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}

\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}

12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

4 0
3 years ago
Other questions:
  • What is true of the mass and volume of all the sinking objects
    5·2 answers
  • The picture shows rays passing through an unknown lens.
    13·2 answers
  • Determine the density of a liquid from the following data .
    13·2 answers
  • If the magnitude of b⃗ increases while its direction remains unchanged, how will the magnetic flux through the coil change?
    7·1 answer
  • An object is 70 micrometer long and 47.66 micrometer wide. How long and wide is the object in km
    8·1 answer
  • Help ME please, I'M BENG TIMED
    5·1 answer
  • Fill in the graph for 50 points
    11·2 answers
  • Which of the following is NOT a characteristic of noble gases? *
    7·1 answer
  • PLZ HELP!! WILL MARK BRAINLIEST!!
    10·1 answer
  • A circuit contains four capacitors in parallel (10 F, 3 F, 7 F, and 1 F). What is the equivalent capacitance of this circuit?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!