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Burka [1]
3 years ago
7

What is the mass of a dragster with a net force of 27,000N and acceleration of 30.0 m/s^2

Physics
1 answer:
garri49 [273]3 years ago
3 0

Answer:

900kg

Explanation:

Data obtained from the question include:

F = 27000N

a = 30.0m/s^2

M =?

F = Ma

M = F/a

M = 27000/30

M = 900kg

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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 is the mass of a pure platinum disk with a volume of 113 cm3? the density of platinum is 21.4 g/cm3
Oduvanchick [21]
V=m×d
m=v/d
m=113/21.4
m=5.28g


6 0
3 years ago
Read 2 more answers
All Physicsts over here plz help in these questions!!!!!!!!!
mrs_skeptik [129]
Hello!

First one we can use that PE=mgh so we have

4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m

Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have

g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2

Hope this helps. Any questions please ask. Thank you.
6 0
3 years ago
Read 2 more answers
Someone help please by providing work and answers please :)
cupoosta [38]
There are two ways to solve this. The longer way is to use those equations to calculate numbers for total distance.

The easier way is to find the area under the graph. That's right, AREA UNDER VELOCITY-TIME graph is the TOTAL DISTANCE travelled!

it's a shortcut.

Let's split up the area into a triangle and rectangle:

Triangle = 0.5(4-0)(10-0) = 20 m
Rectangle = (6-4)(10-0) = 20 m

Total distance = 40 m!
6 0
3 years ago
6. An object moves along the x-axis. The graph shows its position x as a function of time t. Find
raketka [301]

Answer: 1.5 m/s

Explanation:

9-3/5-1= 6/4= 3/2= 1.5

Used x2-x1/t2-t1 to get average velocity from point b to c

7 0
3 years ago
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