At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror's radius of curvature is 0.560 m.A) Locate the image of a patient10.6m from the mirror. B) Indicate whether the image is upright or inverted.C) Determine the magnification of the image.
Answers
Answer:
Explanation:
For a convex mirror, the value of its image distance and its focal length are negative.
using the mirror formula 1/f = 1/u+1/v
f is the focal length = Radius of curvature/2 = 0.560/2
f= 0.28m
u is the object distance = 10.6m
v is the position of the image = ?
On substitution;
1/0.28 = 1/10.6 + 1/-v
3.57 = 0.094 - 1/v
3.57 - 0.094 = -1/v
3.476 = -1/v
v = -1/3.476
v = -0.2877m
B) Since the image distance is negative, this means that the image is an upright and a virtual image. All Upright images has their image distance to be negative.
C) Magnification = Image distance/object distance
Magnification = 0.2877/10.6
Magnification = 0.0271
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Answers
To solve this problem it is necessary to apply the concepts related to the Period based on gravity and length.
Mathematically this concept can be expressed as
Where,
l = Length
g = Gravitational acceleration
First we will find the period that with the characteristics presented can be given on Mars and then we can find the length of the pendulum at the desired time.
The period on Mars with the given length of 0.99396m and the gravity of the moon (approximately will be
For the second question posed, it would be to find the length so that the period is 2 seconds, that is:
Therefore, we can observe also that the shorter distance would be the period compared to the first result given.
Answers
Answer:
0.25( m1m1) , 0.75( m2m2)
Explanation:
Noted the formula for kinetic energy is 1/2(Mass × Velocity).
Therefore the original value of the mass is 0.5, giving half away makes it 0.25 to another mass which is primarily 0.5. This now makes the new mass 0.25+0.5=0.75.
Thank you.
Answers
Answer:
Glycolysis
Explanation:
In human body, glucose levels are regulated by hormones insulin and glucagon, secreted from pancreas. Glucagon from alpha cells and insulin from beta cells of the pancreas. Glucagon are regulated along depending upon the blood sugar levels. During fasting when blood sugar levels are decreased the glucagon levels are increased. Glucagon increases hepatic glucose through glycogenelysis.
So, Glycolysis of glucagon is least likely to be activated by glucagon in hepatocytes
Answers
Answer:
40 m/s due north
Explanation:
Consider that the south direction a negative Y axis and north direction as + Y axis
v1 = 20 m/s South = 20 (-j) m/s
v2 = 20 m/s North = 20 j m/s
Change in velocity = v2 - v1 = 20 j - 20 (-j) = 40 j m/s
So, change in velocity is 40 m/s due north.
Answers
So lets fill out what we have first:
Vi or initial velocity = 20 m/s
Acceleration or a = 4 m/s^2
Time for the motion = 10s
Now, using the four main kinematic equations we can deduce that the best kinematic equation to use in these terms is:
Δx = Vi(t) + 0.5at²
Plug all of our information in:
Δx = (20)(10) + (0.5)(4)(100)
Δx = 400 m
Answer:
400 m
Explanation:
answer on ed
Answers
Answer:
The object's initial temperature is 333.6 K
Explanation:
We first assume that the liquid can only transfer heat to the object through convective heat transfer method.
Let T₀ = the initial temperature of the object
T = temperature of the object at anytime.
The rate of heat transfer from the liquid to the object is given as
Q = -hA (T∞ - T)
T∞ = temperature of the fluid = 400 K
A = Surface area of the object in contact with the liquid = 0.015 m²
h = Convective heat transfer coefficient is given to be = 10 W/(m²K)
The rate of heat gained by the object is given by
mC (d/dt)(T∞ - T)
m = mass of the object = ρV
ρ = density of the object = 100 kg/m³
V = volume of the object = 0.000125 m³
m = ρV = 100 × 0.000125 = 0.0125 kg
C = specific heat capacity of the object = 100 J/(kgK)
The rate of heat loss by the liquid = rate of heat gain by the object
-hA (T∞ - T) = mC (d/dt)(T∞ - T)
(d/dt)(T∞ - T) = - (dT/dt) ( Since T∞ is a constant)
- mC (dT/dt) = -hA (T∞ - T)
(dT/dt) = (hA/mC) (T∞ - T)
Let s = (hA/mC)
(dT/dt) = -s (T - T∞)
dT/(T - T∞) = -sdt
Integrating the left hand side from T₀ (the initial temperature of the object) to T and the right hand side from 0 to t
In [(T - T∞)/(T₀ - T∞)] = -st
(T - T∞)/(T₀ - T∞) = e⁻ˢᵗ
(T - T∞) = (T₀ - T∞)e⁻ˢᵗ
s = (hA/mC) = (10 × 0.015)/(0.0125×100) = 0.12
T = 380 K at t = 10 s
T₀ = ?
T∞ = 400 K
st = 0.12 × 10 = 1.2
(380 - 400) = (T₀ - 400) e⁻¹•²
(-20/0.3012) = (T₀ - 400)
(T₀ - 400) = - 66.4
T₀ = 400 - 66.4 = 333.6 K
Hope this Helps!!!