Answer:
Explanation:
a )
While breaking initial velocity u = 62.5 mph
= 62.5 x 1760 x 3 / (60 x 60 ) ft /s
= 91.66 ft / s
distance trvelled s = 150 ft
v² = u² - 2as
0 = 91.66² - 2 a x 150
a = - 28 ft / s²
b ) While accelerating initial velocity u = 0
distance travelled s = .24 mi
time = 19.3 s
s = ut + 1/2 at²
s is distance travelled in time t with acceleration a ,
.24 = 0 + 1/2 a x 19.3²
a = .001288 mi/s²
= 2.06 m /s²
c )
If distance travelled s = .25 mi
final velocity v = ? a = .001288 mi / s²
v² = u² + 2as
= 0 + 2 x .001288 x .25
= .000644
v = .025 mi / s
= .0025 x 60 x 60 mi / h
= 91.35 mph .
d ) initial velocity u = 59 mph
= 86.53 ft / s
final velocity = 0
acceleration = - 28 ft /s²
v = u - at
0 = 86.53 - 28 t
t = 3 sec approx .
Answer:
The magnitude of the torque is 263.5 N.
Explanation:
Given that,
Applied force = 31 N
Distance from the axis = 8.5 m
She applies her force perpendicularly to a line drawn from the axis of rotation
So, The angle is 90°
We need to calculate the torque
Using formula of torque
Where, F = force
d = distance
Put the value into the formula
Hence, The magnitude of the torque is 263.5 N.
Answer/Explanation:
The weight of an object is defined as the force that is exerted due to the gravitational force.
Mathematically, it can be written as :
W = m g
Where
m is the mass of the object
g is the acceleration due to gravity
Also,
We know that the value of g varies with respect to the location. At the equator, the value of g is less as compared to the poles.
The feature of an object that affects its weight are :
Mass of the object
Location of the object
How much force Earth exerts on the object
Answer:
a) r = 6122 m and b) v = 32.5 m / s
Explanation:
a) The train in the curve is subject to centripetal acceleration
a = v2 / r
Where v is The speed and r the radius of the curve
They indicate that the maximum acceleration of the person is 0.060g,
a = 0.060 g
a = 0.060 9.8
a = 0.588 m /s²
Let's calculate the radius
v = 216 km / h (1000m / 1km) (1 h / 3600 s =
v = 60 m / s
r = v² / a
r = 60² /0.588
r = 6122 m
b) Let's calculate the speed, for a radius curve 1.80 km = 1800 m
v = √a r
v = √( 0.588 1800)
v = 32.5 m / s
