Answer:
distance changing at rate of 3.94 inches/sec
Explanation:
Given data
wall decreasing at a rate = 9 inches per second
ladder L = 152 inches
distance h = 61 inches
to find out
how fast is the distance changing
solution
we know that
h² + b² = L² ..................1
h² + b² = 152²
Apply here derivative w.r.t. time
2h dh/dt + 2b db/dt = 0
h dh/dt + b db/dt = 0
db/dt = - h/b × dh/dt .............2
and
we know
h = 61
so h² + b² = L²
61² + b² = 152²
b² = 19383
so b = 139.223
and we know dh/dt = -9 inch/sec
so from equation 2
db/dt = -61/139.223 (-9)
so
db/dt = 3.94 inches/sec
distance changing at rate of 3.94 inches/sec
Answer:
Terminal velocity of object = 12.58 m/s
Explanation:
We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.
Gravitational force = mg = 80 * 9.8 = 784 N
Drag force = 
Equating both, we have

So v = 12.58 m/s or v = -15.58 m/s ( not possible)
So terminal velocity of object = 12.58 m/s
Answer:
455165.278 m
Explanation:
P = Power = 3.7 W
v = Velocity = 10.7 m/s
Amount of fat = 4 g
1 gram of fat provides about 9.40 (food) Calories
Energy given by 4 g of fat

Time required to burn the fat

Distance traveled by the bird

The bird will fly 455165.278 m
<span>Answer:The weight of the door creates a CCW torque given by
Tccw = 145 N*3.13 m / 2
You need a CW torque that's equal to that
Tcw = F*2.5 m*sin20</span>
Answer: SG = 2.67
Specific gravity of the sand is 2.67
Explanation:
Specific gravity = density of material/density of water
Given;
Mass of sand m = 100g
Volume of sand = volume of water displaced
Vs = 537.5cm^3 - 500 cm^3
Vs = 37.5cm^3
Density of sand = m/Vs = 100g/37.5 cm^3
Ds = 2.67g/cm^3
Density of water Dw = 1.00 g/cm^3
Therefore, the specific gravity of sand is
SG = Ds/Dw
SG = (2.67g/cm^3)/(1.00g/cm^3)
SG = 2.67
Specific gravity of the sand is 2.67