Q1.

Physics

1 answer:

Leno4ka [110]3 years ago

4 0

A) force = mass • acceleration
Convert: 1.870 KN -> 1870 N
1870N = mass • 1.83 m/s^2
———- —————-
1.83 m/s^2. 1.83 m/s^2
-> approx. 1,021.9 grams :D

b) Vi = Vf - at
0 m/s = Vf - 1.83 m/s^2 • 16s
0m/s = Vf - 29.28
+29.28 + 29.28
Vf = 29.28 m/s :D

c) x = Vi + 1/2 • at^2
A to B
T= 15s
Vi = 16m / s
A = 1.83m/s^2
x= 16 + 1/2 • 1.83(15)^2)
x = approx. 221. 9 meters

C to D
T = 9s
Vi= 30m/s
x = 30 + 1/2 1.83(9)^2
x= approx. 104.1 meters

Therefore, when the car is moving at a CONSTANT RATE, it’s traveling farther than when slowing down cause it travel 221.9 meters while slowing down, it’s going only 104.1 meters.

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A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$