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riadik2000 [5.3K]
3 years ago
15

Could a nucleus that has one proton but no neutrons exist?

Physics
1 answer:
diamong [38]3 years ago
8 0
Yes, but there is only 1 atom like that and is is hydrogen. Hydrogen is the only element that could have a nucleus with one proton and no neutrons exist.
You might be interested in
Anika asks Eva to roll a basketball and then a bowling
Alex73 [517]

Answer:

the bowling ball, because it has more mass and  therefore more inertia

Explanation:

As per law of inertia we know that if an object is having more inertia then it is difficult to change state of motion.

Inertia is the property of mass of an object which always resist to change the state of motion of the object.

If an object has more inertia then it is more difficult to change the state of motion.

Now we know that we have one bowling ball and one basket ball, since bowling ball is having more mass then it must have more inertia so it is difficult to start the motion in bowling ball.

So correct answer will be

the bowling ball, because it has more mass and  therefore more inertia

8 0
3 years ago
Javier wants to improve his flexibility and increase his personal discipline. Which community resource should he select?
Dovator [93]
Karate class, kicking high and ducking fast makes you much more flexible and you must pay respects to those ou fight with and be intune with yourself to do it right, in other words karate is the answer.
7 0
3 years ago
____ is a type of circuit were there is a direct connection between two points in a circuit that aren't supposed to be directly
PtichkaEL [24]

Answer:

Short

Explanation:

I took the same test

7 0
3 years ago
A wave traveling in water has a frequency of 500.0 Hz and a wavelength of 3.00 m. What is the speed of the wave?
Sergeu [11.5K]

Answer:

1500 m/s

Explanation:

Recall that for a wave,

Speed = frequency x wavelength

here we are given frequency = 500 Hz and wavelength = 3m

simply substitute into above equation

Speed = 500 Hz x 3m

= 1500 m/s

6 0
3 years ago
A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

K=\frac{1}{2}mv^2

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

4 0
3 years ago
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