Determine the average speed of a runner who runs 15 Km in one hour and 30 minutes

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

What is endurance? a). the ability to run faster b). a combination of balance and coordination c). how much you can stretch d).

Alex787 [66]

Answer:

D. the ability to exercise for longer periods of time

Explanation:

For example, when someone does endurance training, they are stretching their body's ability to do a certain exercise for longer times as opposed to increasing strength.

8 0

3 years ago

Help Please

densk [106]

Answer:

Energy=3.1times 10^-17 J

Rest mass: 6.2 kg

Speed: 47.5 m/s

Wavelength: 2.659 times 10^-6

Momentum: 67.3 kg(m/s)

Explanation:

3 0

3 years ago

At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th

Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r

So (86kg* 101^2)/(167) =

Fc=5253 N

4 0

3 years ago

Three collinear forces,F1=45N west,F2=63N east and an unknown force F3 are applied to an object.The resultant force of the three

Greeley [361]

Take east to be the positive direction. Then the resultant force from adding F₁ and F₂ is

F₁ + F₂ = (-45 N) + 63 N = 18 N

which is positive, so it's directed east.

To this we add a third force F₃ such that the resultant is 12 N pointing west, making it negative, so that

18 N + F₃ = -12 N

F₃ = -30 N

So F₃ has a magnitude of 30 N and points west.

6 0

3 years ago