Answer:
Time - taken = 2.5 s
deceleration= -8 m/s²
Solution:
Given:
speed, v = 8 m/s
distance, d = 20m
To Find:
deacceleration = ?
As we know speed is defined as
v = d/t
plugging in the values
t = 20/ 8
t = 2.5s
Now from deceleration formula
a = - v/ t
a = - 20/ 2.5
a = - 8 m/s²
Thus, the time taken and acceleration is 2.5 s and -8 m/s²
respectively.
Learn more about deceleration here:
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<span>D. Along two converging oceanic plate boundaries</span>
Answer:
343/1500
Explanation:
Power: This can be defined as the product force and velocity. The S.I unit of power is Watt (w).
From the question,
P' = mg×v................. Equation 1
Where P' = power used to gain an altitude, m = mass of the engine, g = acceleration due to gravity of the engine, v = velocity of the engine.
Given: m = 700 kg, v = 2.5 m/s, g = 9.8 m/s²
Substitute into equation 1
P' = 700(2.5)(9.8)
P' = 17150 W.
If the full power generated by the engine = 75000 W
The fraction of the engine power used to make the climb = 17150/75000
= 343/1500
Answer: v= 160ft/s
a=32ft/s^2 constant
Explanation:
s(t)=400-16t^2 derivative of position is velocity v(t) and derivative of velocity is acceleration a(t) so let s(t)=0 to find the time of flight to reach the ground and take the two derivatives and use the time found and solve. Also acceleration is a constant as it’s gravity.
0=400-16t^2
400=16t^2
25=t^2
t=5s
ds/dt=v(t)=0-32t
dv/dt=a(t)=-32 constant(gravity)
v(t)=-32(5s)= -160ft/s negative sign is only showing direction
Answer:
Explanation:
Image of distant object will be made at far point or at 52.5 so
object distance u = infinity
image distance v = - 52.5 cm
focal length required = f
Lens formula
1 / v - 1 / u = 1 / f
1 / - 52.5 - 0 = 1 / f
f = -52.5 cm
= -.525 m
Power P = 1 / f = - 1 / .525
= - 1.90
now , for eye with glass we shall find new near point .
v = ?
u = - 17.2 cm
f = - 52.5 cm
1 / v - 1 / u = 1 / f
1 / v + 1 / 17.2 = - 1 / 52.5
1 / v = - 1 / 17.2 - 1 / 52.5
= - .05813 - .019
= - .07713
u = - 12.96 cm
so new near point will be 12.96 cm