Kinetic energy is energy that is in motion, thats all I remember
Answer:
The second projectile was 1.41 times faster than the first.
Explanation:
In the ballistic pendulum experiment, the speed (v) of the projectile is given by:
<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum. </em>
To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:
(1)
<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively. </em>
Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):
Therefore, the second projectile was 1.41 times faster than the first.
I hope it helps you!
The more arms it has the less of a chance the prey has to swim away.
Answer:
the new resister is 11 ohms.
Explanation:
Set it up like this.
1/x + 1/1.1 = 1 Subtract 1/1.1 from both sides
1/x = 1 - 1/1.1
1 - 1/1.1 = 1/11
1/x = 1/11 Cross multiply
11 = x
If 1/11 bothers you, you could do it it another way.
1 - 1/1.1 = (1.1 - 1 ) / 1.1 = 0.1 / 1.1 Multiply top and bottom by 10
0.1*10/(1.1 * 10 ) = 1 / 11
Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
