What is the mass of a 50 kg person on earth?

How does an atom of potassium-41 become a potassium ion with a +1 charge? 19 K 39.10

stiv31 [10]

It is very difficult for an atom to accept a proton. It can only be done under very special circumstances. So A and C are both incorrect. I don't see how D is possible. The atom does lose 1 electron, but how it gets 21 is think air.

The answer is B which is exactly what happens.

5 0

3 years ago

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A 14.3-g bullet is fired into a 5.21 kg block of wood. The block is attached to a spring that has a spring force constant of 450

Natasha2012 [34]

Answer:

The initial speed of the bullet is $v_{o} = 889.199\,\frac{m}{s}$ .

Explanation:

The collision between bullet and block is inelastic and let suppose that motion occurs on a horizontal surface, so that changes in gravitational potential energy can be neglected. Initially, the intial speed of the bullet-block system can be determined with the help of the Work-Energy Theorem and the Principle of Energy Conservation:

$K = U_{k} + W_{loss}$

$\frac{1}{2}\cdot (5.224\,kg)\cdot v^{2} = \frac{1}{2}\cdot \left(450\,\frac{N}{m}\right)\cdot (0.22\,m)^{2} + (0.35)\cdot (5.224\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right) \cdot (0.22\,m)$

The initial speed of the bullet-block system is:

$v \approx 2.383\,\frac{m}{s}$

Now, the initial speed of the bullet is determined by applying the Principle of Momentum Conservation:

$(0.014\,kg)\cdot v_{o} = (5.224\,kg)\cdot \left(2.383\,\frac{m}{s} \right)$

$v_{o} = 889.199\,\frac{m}{s}$

6 0

3 years ago

If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the c

Lemur [1.5K]

Answer:

$\mu_s \geq 0.27$

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

$f_s=\frac{mv^{2}}{R}$

But we know that:

$f_s\leq \mu_s N$

And the normal force is given by the sum of the forces in the vertical direction:

$N-mg=0 \implies N=mg$

Finally, we have:

$f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27$