Answer:
reduced performance due to stereotype threat
Explanation:
Answer:
u = - 38.85 m/s^-1
Explanation:
given data:
acceleration = 2.10*10^4 m/s^2
time = 1.85*10^{-3} s
final velocity = 0 m/s
from equation of motion we have following relation
v = u +at
0 = u + 2.10*10^4 *1.85*10^{-3}
0 = u + (21 *1.85)
0 = u + 38.85
u = - 38.85 m/s^-1
negative sign indicate that the ball bounce in opposite directon
Answer
given,
mass of the ball = 3 kg
swing in vertical circle with radius = 2 m
work done by the gravity = ?
work done by the tension = ?
Work done by the gravity = - m g Δh
Δ h = 2 + 2 = 4 m
Work done by the gravity =
= -117.6 J
work done by gravity is equal to -117.6 J
Work done by tension will be equal to zero.
Zero because tension is always perpendicular to velocity
work done by tension is equal to 0 J
Answer:
a) 17.33 V/m
b) 6308 m/s
Explanation:
We start by using equation of motion
s = ut + 1/2at², where
s = 1.2 cm = 0.012 m
u = 0 m/s
t = 3.8*10^-6 s, so that
0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²
0.012 = 0.5 * a * 1.444*10^-11
a = 0.012 / 7.22*10^-12
a = 1.66*10^9 m/s²
If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where
E = electric field
m = mass of proton
a = acceleration
q = charge of proton
E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19
E = 2.77*10^-18 / 1.6*10^-19
E = 17.33 V/m
Final speed of the proton can be gotten by using
v = u + at
v = 0 + 1.66*10^9 * 3.8*10^-6
v = 6308 m/s
The available options are:
Mint is a dicot.
Mint is a monocot.
Mint is an angiosperm.
Mint is a bulb plant.
Answer:
Mint is a dicot.
Explanation:
Given the fact that Mint is considered to be a member of Lamiaceae, an angiosperm plant which is characterized by typically having leaves that consist of reticulate vacation and appears like veins in structure. It also has a seed that contains two cotyledons.
Hence, it is considered a DICOT PLANT due to these characteristics. The botanical name of Mint is referred to as Mentha arvensis.