Answer:
99
Explanation:
First strike=25 in
Second strike=4/5*25=20 in
Third strike=4/5*20=16 in
Fourth strike=4/5*16=12.8 in
Fifth strike=4/5*12.8=10.24 in
sixth strike=4/5*10.24=8.192 in
seventh strike=4/5*8.192=6.5536 in
Total=(25+20+16+12.8+10.24+8.192+6.5536) in=98.7856 in
Taking this as a geometric series
where a is initial value taken as 25 for this case and r is rate taken as 4/5 or 0.8 in our case
To the nearest inches, sum is 99 in
Answer:
The fastest object is the sphere, so it is the winner
Explanation:
To know which object will arrive faster down, let's look for the velocity of the center of mass of each object. Let's use the concept of mechanical energy
Highest point
Em₀ = U = mg y
Lowest point
= K =
+
= ½ I w² + ½ m
²
Angular velocity is related to linear velocity.
v = w r
w = v / r
= ½ I
²/r² + ½ m
²
= ½ (I / r² + m)
²
Energy is conserved
Em₀ =
mg y = ½ (I / r² + m) ²
= √2 g y / (I / mr² +1)
With this expression we can know which object arrives as a higher speed, therefore invests less time and is the winner. Let's calculate the speed of the center of mass of each
Ring
I = m r²
= √ (2 g y / (m r² / mr² + 1))
= √ (2gy 1/2)
= (√ 2gy) 0.707
Solid sphere
I = 2/5 m r²
= √ (2gy / (2/5 m r² / mr² + 1)
= √ (2gy / (7/5))
= √ (2gy 5/7)
= (√ 2gy) 0.845
Cylinder
I = ½ m r²
= √ (2gy / ½ mr² / mr² + 1)
= √ (2gy / (3/2))
= √ (2g y 2/3)
= (√ 2gy) 0.816
The fastest object is the sphere, so it is the winner when descending the ramp
In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?
Answer:
9.1Ω
Explanation:
The circuit diagram has been attached to this response.
(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e
=> ------------(i)
From the question;
R1 = 3Ω,
R2 = 7Ω
Substitute these values into equation (i) as follows;
Ω
(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.
Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e
R = Rₓ + R3
Rₓ = 2.1Ω
R3 = 7Ω
=> R = 2.1Ω + 7Ω = 9.1Ω
Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω
