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your target heart rate is between 64% and 76% of your maximum heart rate
We know that a charge moving in a magnetic field is subject to the force:
F = q · v · B
But we also know that:
F = m · a
Therefore, it must be:
m · a = <span>q · v · B
And solving for a:
</span>a = <span>q · v · B / m
Recall that for a proton:
q = 1.6</span>×10⁻¹⁹ C
m = 1.673×10⁻²⁷ kg
Now, you can find:
a = 1.6×10⁻¹⁹ · 7.0 · 1.7 / <span>1.673×10⁻²⁷
= 1.14</span>×10⁹ m/s²
Hence, the acceleration of the proton is 1.14<span>×10⁹ m/s²</span>.
Answer:
1/3
Explanation:
We can solve the problem by using the lens equation:

where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
Here we have a divering lens, so the focal length must be taken as negative (-f). Moreover, we know that the object is placed at a distance of twice the focal length, so

So we can find q from the equation:

Now we can find the magnification of the image, given by:

Resistance of our body is given as

voltage applied across the body is

now by ohm's law current pass through our body is given by

![i = \frac{120}{30,000}[\tex][tex]i = 4 * 10^{-3} A](https://tex.z-dn.net/?f=i%20%3D%20%5Cfrac%7B120%7D%7B30%2C000%7D%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5Di%20%3D%204%20%2A%2010%5E%7B-3%7D%20A)
So current from our body will be 4 * 10^-3 A
Answer:
768000 joule
Explanation:
As we know that
K.E = 0.5 x mv2
Here m stands for mass, the measure of how much matter is in an object, and v stands for the velocity of the object, or the rate at which the object changes its position.
putting the value we get
K.E = 0.5 x (1500KG)(32 m/s)2
K.E =768000 J