Answer:
The sled needed a distance of 92.22 m and a time of 1.40 s to stop.
Explanation:
The relationship between velocities and time is described by this equation: 
, where 
is the final velocity, 
is the initial velocity, 
the acceleration, and 
is the time during such acceleration is applied.
Solving the equation for the time, and applying to the case: 
, where 
because the sled is totally stopped, 
is the velocity of the sled before braking and, 
is negative because the deceleration applied by the brakes.
In the other hand, the equation that describes the distance in term of velocities and acceleration:
, where 
is the distance traveled, is the initial velocity, the time of the process and, is the acceleration of the process.
Then for this case the relationship becomes: 
.
<u>Note that the acceleration is negative because is a braking process.</u>
Answer:
x component 60.85 m
y component 101.031 m
Explanation:
We have given distance r = 118 km
Angle which makes from ground = 58.9°
(a) X component of distance is given by 
(b) Y component of distance is given by 
These are the x and y component of position vector
Answer:
P = 17.28*10⁶ N
Explanation:
Given
L = 250 mm = 0.25 m
a = 0.54 m
b = 0.40 m
E = 95 GPa = 95*10⁹ Pa
σmax = 80 MPa = 80*10⁶ Pa
ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m
We get A as follows:
A = a*b = (0.54 m)*(0.40 m) = 0.216 m²
then, we apply the formula
ΔL = P*L/(A*E) ⇒ P = ΔL*A*E/L
⇒ P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)
⇒ P = 24624000 N = 24.624*10⁶ N
Now we can use the equation
σ = P/A
⇒ σ = (24624000 N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa
So σ > σmax we use σmax
⇒ P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N
Force is defined as the rate of change of momentum.
The initial amount of momentum is
because water stops when it hit the wall total change of momentum must be
.
Now let's calculate the force.
We need to find
. This is the amount of water hiting the wall per second.
Our final formula would be:
And now we can calculate the answer:
Answer:-q
Explanation:
Given
Capacitor is charged to a battery and capacitor acquired a charge of q i.e.
+q on Positive Plate and -q on negative Plate.
If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by
where A=area of capacitor plate
d=Separation between plates
This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q
