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Specific heat is the amount of heat required to raise the temperature of a unit substance!. The most common example!. The specific heat of water is 1 !. And, lot of plays and animals live under the water and if the specific heat of the water would be low, then the aquatic organisms would die, but the specific heat is high and it is hard for the sun to increase the temperature!.
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Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original so it has 8.8 cm with the same orientation as original and it is a virtual imagen.
Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:
1/p+1/q=1/f where p and q represents the distance to the mirror for the object and imagen, respectively. f is the focal length for the concave mirror.
replacing the values we obtain:
1/8.3+1/q=1/15.2
so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3
then q=-18.28 cm
The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2
We also add a picture to see the imagen formation for this case.
Answer:
Wavelength is 471 nm
Explanation:
Given that,
Lines per unit length of diffraction grating is 500 lines/mm.
The third maxima from the central maxima (m=3) is at an angle of 45°
We need to find the color of laser light shines through a diffraction grating.
The condition for maxima is :
d = 1/N, N = number of lines per mm
Answer:
the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
Explanation:
Given that;
final velocity v = 0
initial velocity u = 15m/s
time taken t = 4 s
acceleration a = ?
from the equation of motion
v = u + at
we substitute
0 = 15 + a × 4
acceleration a = -15/4 = - 3.75 m/s²
the negative sign tells us that its a deacceleration so the sign can be ignored.
Deacceleration due to friction a = μ × g
we substitute
3.75 = μ × 9.8
μ = 3.75 / 9.8 = 0.3826 ≈ 0.38
Therefore the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
Answer:
Explanation:
Given
Maximum height H = 300m
Range (horizontal distance) = 380m
Required
Initial speed U and the angle of the ball when it was launched.
Range = U√2H/g
380 = U√2(300)/9.8
380 = U√600/9.8
380 = 7.8246U
U = 380/7.8246
U = 48.57m/s
The initial speed is 48.57m/s
b) Using the formula for calculating time of flight;
T = 2Usin theta/g
9 = 2(48.57)sin theta/9.8
9*9.8 = 97.14sin theta
88.2 = 97.14sin theta
88.2/97.14 = sin theta
sin theta = 0.9079
theta = sin^-1(0.9079)
theta = 65.23°
hence the angle when the ball was launched is 65.23°
