A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first 1/4 of that distance, its acceleration is +6.25 m/s2. Through the next 3/4 of that distance, its acceleration is -2.08 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed?
<span>Solve for the time at the 1/4 mark. That's 225 m. How? d = (1/2)at^2 ( initial velocity zero). Thus 225 = (1/2) 6.25 t^2. t^2 = ( 225 * 2 ) / 6.25. t = 8.5 sec. </span>
<span>At the other end t^2 = (675 * 2) / 2.08 -- we reversed the sign and ran time backwards. t = 25.5 sec. </span>
<span>So total time is 8.5 + 25.5 or 34 sec. </span>
<span>Since zero initial velocity: v^2 = 2 a d. Here, v^2 = 2 * 6.25 * 225. v = 53 m/s. That's the fastest speed since braking then occurs.</span>
Answer:
Explanation:
i. CW moment = 10 N (10 cm) + 30 N (30 cm) - 60 N (40 cm) = - 1400 N-cm
ii. ACW momenet = 60 N (40 cm) - 10 N (10 cm) + 30 N (30 cm) = 1400 N-cm
iii. No. The lever is not balanced in the situation. Because the moment is ± 1400 N-cm. if balance, the moment must be Zero.
iv. the location of 10N by keeping the other loads unchanged to balance the lever is 150 cm
take moment from Δ (support)
60(40) = 10(x) + 30(30)
2400 = 10x + 900
10x = 2400 - 900
10x = 1500
x = 1500/10
x = 150 cm
therefore, the location of 10N by keeping the other loads unchanged to balance the lever is 150 cm
Answer:
Explanation:
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Answer:
The capacitor having less distance of separation has a stronger electric field.
Explanation:
The capacitors are identical and only difference between them is that one has twice the plate separation of the other. Therefore, capacitance of the given capacitors C1 and C2 is,
C1= Aε/d and C2=Aε/2d
The charges Q1 and Q2 on the capacitors of capacitance C1 and C2 respectively, is then given by the equation,
Q1=VC1
Q1=VAε/d
Q2=VC2
Q2=VAε/2d
Therefore, the surface charge density σ1 and σ2 for the capacitors is,
σ1=Q1/A
σ1=VAε/(d*A)
σ1=Vε/d
Similarly,
σ2=Q2/A
σ2=Vε/2d
The electric field between the plates is directly proportional to the surface charge density. And so electric field is inversely proportional to the distance of separation. Therefore the capacitor whose distance of separation is less has a stronger electric field.
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Answer:
The car will travel a distance of 17.45 meters.
Explanation:
Given:
Initial velocity 
= 0
Final velocity 
= 7.6 m/s
Time taken = 4.6 s
Acceleration = (Final velocity - Initial Velocity )/time
We have to calculate total distance traveled by the car.
Let the distance traveled be 'd'
Equation of motion:
Plugging the values.
⇒
⇒
⇒
The car will travel a distance of 17.45 meters for the above case.
