Refer to the diagram shown below.
By definition momentum = mass * velocity.
Before throwing the ball:
The initial momentum is
P₁ = 0.
After throwing the ball:
Let u = the backward velocity of the quarterback.
The momentum is
P₂ = (0.43 kg)*(15 m/s) + (80 kg)*(- u m/s)
Conservation of momentum requires that
P₂ = P₁
6.45 - 80u = 0
u = 6.45/80 = 0.0806 m/s
Answer: 0.08 m/s backward
Answer:5.35nm
Explanation:
Consider that 1 inch is = 0.0254m
we have,
1m= 1x10^9 nm
While:
0.0254m = 2.54x10^7nm
1/55 (2.54x10^7) = 4.6181 x 10^5nm
1 day= 24 hrs
= (24x60) when calculating in min
= (24x60x60) calculating in seconds we have:
= 8.64x10⁴sec
In 8.64x10^4 seconds, the hair grows by 4.6181 x 10^5nm
Therefore, the amount by which the hair grows in 1 second will be;
= (4.6181 x 10^5)/(8.64x10^4)
= 5.35nm
The rate of growth will be 5.35nm
Answer:
The Ferris wheel's tangential (linear) velocity if the net centripetal force on the woman is 115 N is <u>3.92 m/s</u>.
Explanation:
Let's use <u>Newton's 2nd Law</u> to help solve this problem.
The force acting on the Ferris wheel is the centripetal force, given in the problem: .
The mass "m" is the <u>sum</u> of the man and woman's masses: .
The acceleration is the centripetal acceleration of the Ferris wheel: .
Let's write an equation and solve for "v", the tangential (linear) acceleration.
The Ferris wheel's tangential velocity is 3.92 m/s.
Answer: the electric field is equal to: E=0 for r<2 cm; E= -2.29*10^4/r^2 for 2cm<r<4cm; E=0 for 4cm<r<5 cm; E=5.26*10^4/r^2 for r>5 cm
the electric field in N/C units
Explanation: In order to find the electric field for all r values, we have to use the definition of electric field and Gaussian law.
In this sense, for r<2 cm as it is inside a conductor teh electric field is zero.
for 2cm< r< 4 cm we applied the field from a spherical charge distribution so by the Gaussian law we find the total charge inside the gaussian surface so
E.4π r^2= Q inside/ε0 = -2.55μC/ε0
Idem for other regions.
for 4 cm<r< 5 cm in the outsider conductor the E=0
Finally, for r>5 cm
E.4π r^2=Q inside/ε0=(8.40-2.55)μC/ε0
Solution:
Mass(m)=20kg
Displacement (s)=1.5m
For 1 kg force of gravity=10N
We have,
F=mg
or,10=1xg
or g=10m/s^2
Then,
Total work done by the coolie(W)=force x displacement
or,W=mxgxs
=20x10x1.5
=300J