Question: Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.
Answer:
1.29 m/s²
Explanation:
From the question,
a = (v-u)/t............................ Equation 1
Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
Given: v = 13 m/s, u = 35 m/s, t = 17 s.
a = (13-35)/17
a = -22/17
a = -1.29 m/s²
Hence the deceleration of the car is 1.29 m/s²
The watts determine the brightness. Watt is the unit of Power. And Power is equal to Voltage x Amps (current). So both the current and voltage determine the brightness.
Answer:
a) 1.57, 1.61
B 1.91x10^8m/s , 1.86 x 10^8m/s
Explanation:
Ok we know that refractive index is given as = sin i/ sin r
And i is the angle of incidence in the air and r is angle of refraction in the glass,
So we can sa y
for red light n = sin75/sin38.1 = 1.57
and for violet, n =sin75/sin36.7 = 1.61
So
(a) 1.57, 1.61
(b) we know that
Refractive index n = c/v,
where c is speed off light in air and v is speed of light in glass,
So , v = c/n
for red light v = c/1.57 = 1.91 x 10^8 m/s
and for violet light, v = c/1.61 =
1.86 x 10^8 m/s
angular vel to tangential vel
v=r omega
v = 56 x 100/60 x 2 pi
v = 56x5/3x6
v=560m/s as estimate
100 revs, 5.00m
Explanation:
A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image be located, and how tall will it be? Please show all work. Use the steps below
Step 1: Find the focal length, step 2: find the image distance using the mirror equation step 3: Use the magnification relationship to find the image height.