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hichkok12 [17]
2 years ago
12

What are the advantages of strength exercises?

Physics
1 answer:
Sever21 [200]2 years ago
3 0

Answer: Well some advantages of strength is the fact that: you have a stronger heart, Immune system, more energy, life span, stamina, positive thoughts, and also believe it or not. An easier connection with the deeper part of earth

Explanation:

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You have two contentious friends, Chris and Pat, and you’ve quickly discovered that they need you to resolve arguments they have
pshichka [43]

Answer:

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points,

Explanation:

To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

             v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.

Therefore neither of them is right.

3 0
3 years ago
Why are the wheels on a roller coaster hot after a ride. (I know it’s due to friction but I need more of a detailed answer).
telo118 [61]
2 someone's glasses could fall of because of the inertia
roller coaster is in motion but your glasses are opposing and resisting that motion due to inertia....

1 there is a possibility that electrons move from wheels to the rail and vice versa which also means that it is producing some kind of electrical energy

3 question can't answer
8 0
3 years ago
Which of the following is a type of T cell?
Bess [88]

Answer:

D

Explanation:

7 0
2 years ago
Read 2 more answers
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
NEED ASAP PLEASE HELP
sveta [45]
1. GPE - 40 * 2 * 10 = 800j
3 0
3 years ago
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