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Valentin
3 months ago
12

The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un

its. What is the instantaneous acceleration of the object when t = 4.1 s?
Physics
1 answer:
serg [3.5K]3 months ago
6 0

Answer:

The response to your inquiry is: 15 m/s²

Explanation:

Equation    x = at³ - bt² + ct

a = 4.1 m/s³

b = 2.2 m/s²

c = 1.7 m/s

First we calculate x at t = 4.1 s

x = 4.1(4.1)³ - 2.2(4.1)² + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we calculate speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s²

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If the axes of the two cylinders are parallel, but displaced from each other by a distance d, determine the resulting electric f
serg [3582]

Response:

E =  ρ ( R1²) / 2 ∈o R

Clarification:

Provided information

Two cylinders are aligned parallel

Distance = d

Radial distance = R

d < (R2−R1)

To determine

Express the response using the variables ρE, R1, R2, R3, d, R, and constants

Solution

We have two parallel cylinders

therefore, area equals 2 \pi R × l

And we apply Gauss's Law

EA = Q(enclosed) / ∈o......1

Initially, we calculate Q(enclosed) = ρ Volume

Q(enclosed) = ρ ( \pi R1² × l )

Thus, inserting all values into equation 1

produces

EA = Q(enclosed) / ∈o

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This simplifies to

E =  ρ ( R1²) / 2 ∈o R

6 0
2 months ago
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