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Vsevolod [243]
3 years ago
9

Tafu is working with subatomic particles in the physics lab. a positron is traveling in a straight line down the particle accele

rator. tafu models its position with the function p(t)= t2 t â 1 where t>1. in the limit calculations below, if you need to use -â or â, enter -infinity or infinity. (a) calculate the limit
Physics
1 answer:
KatRina [158]3 years ago
4 0

Answer:

Infinity

Explanation:

When you have a limit approaching infinity one method you can use is to remember that the highest power dominants all of the other ones so when we have (t^2)/t-1 like in your problem, we see that t^2 has the highest power of 2 so if that's going to infinity and it's not being divided by any powers higher than it will just be inifinity.

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bixtya [17]

Answer:

AS- X 3.42 Y 3. B) X Y c) x Ross TET V V. a 1.71 1.71 LLL LLL 2.42 N al

Explanation:

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3 years ago
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How should you plan to excercise if you want to remain healthy past the age of 65
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Run,karate,aerobic exercise
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2 years ago
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When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 2.5 x 10-15 m before
Tasya [4]

Explanation:

It is given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, d=2.5\times 10^{-15}\ m

Speed of light, c=3\times 10^{8}\ m\s

Let t is the time interval required for the strong interaction to occur. The speed is given by :

c=\dfrac{d}{t}

t=\dfrac{d}{c}

t=\dfrac{2.5\times 10^{-15}\ m}{3\times 10^{8}\ m/s}

t=8.33\times 10^{-24}\ s

So, the time interval required for the strong interaction to occur is 8.33\times 10^{-24}\ s. Hence, this is the required solution.

8 0
3 years ago
A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a d
igor_vitrenko [27]

Answer:

Explanation:

potential energy of compressed spring

= 1/2 k d²

= 1/2 x 730 d²

= 365 d²

This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .

Kinetic energy after crossing the rough patch

= 1/2 x 1.2 x 2.3²

= 3.174 J

Loss of energy

= 365 d² - 3.174  

This loss is due to negative work done by frictional force

work done by friction = friction force x width of patch

= μmg d ,   μ = coefficient of friction , m is mass of block , d is width of patch

= .44 x 1.2 x 9.8 x .05

= .2587 J

365 d² - 3.174   = .2587

365 d² = 3.4327

d² = 3.4327 / 365

= .0094

d = .097 m

= 9.7 cm

If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .

5 0
3 years ago
You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo
AlekseyPX

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

(c) When same force is applied to the shopping cart on a wet surface, it will move faster.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

  • F is the applied force
  • Ff is the frictional force

ma = F - Ff

where;

  • a is acceleration of the cart
  • m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net)  = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

Learn more about frictional force here: brainly.com/question/24386803

#SPJ1

4 0
2 years ago
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