Answer:
In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.
Answer:
Efficiency = StartFraction T Subscript h Baseline minus T Subscript C Baseline over T Subscript h Baseline EndFraction times 100. Efficiency equals T Subscript c Baseline minus T Subscript h Baseline over T Subscript h Baseline all times 100.
Answer:
a= - 6.667 m/s²
Explanation:
Given that
The initial speed of the box ,u= 20 m/s
The final speed of the box ,v= 0 m/s
The distance cover by box ,s= 30 m
Lets take the acceleration of the box = a
We know that
v²= u ² + 2 a s
Now by putting the values in the above equation we get
0²=20² + 2 a x 30
a= - 6.667 m/s²
Negative sign indicates that velocity and acceleration are in opposite direction.
Therefore the acceleration of the box will be - 6.667 m/s² .
Answer:
71.85 m/s
Explanation:
Given the following :
Length of skid marks left by jaguar (s) = 290 m
Skidding Acceleration (a) = - 8.90m/s²
Final velocity of jaguar (v) = 0
Speed of Jaguar before it Began to skid =?
Hence, initial speed of jaguar could be obtained using the formula :
v² = u² + 2as
Where
v = final speed of jaguar ; u = initial speed of jaguar(before it Began to skid) ; a = acceleration of jaguar ; s = distance /length of skid marks left by jaguar
0² = u² + (2 × (-8.90) × 290)
0 = u² + (-5,162)
u² = 5162
Take the square root of both sides
u = √5162
u = 71.847 m/s
u = 71.85m/s
Answer:
53.64 m/s
Explanation:
Applying,
a = (v-u)/t............. Equation 1
Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
make u the subject of the equation
u = v-at............. Equation 2
From the question,
Given: a = -12 mph/s = -5.364 m/s², t = 10 seconds, v = 0 m/s (comes to stop)
Substitute these values into equation 2
u = 0-(-5.364×10)
u = 0+53.64
u = 53.64 m/s
