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3 years ago

What is the acceleration of a car that travels in a straight line?

2 answers:

6 0

Acceleration occurs when there is a change in speed or direction. If it travels in a straight line, there is no speed or change in direction as it is constant througout, hence 0 acceleration.

hope this helps!! ✨

Bond [772]3 years ago

4 0

<span>Average and instantaneous (by contrast there is also centripetal for angular motion)

</span>

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một ô tô chuyển động nhanh dần đều.Sau 10s vận tốc của xe tăng từ 2m/s lên 6m/s.Tính gia tốc của ô tô

Serjik [45]

Answer:

áp dụng công thức í, mình thấy câu này có rắc rối gì đâu

8 0

3 years ago

You leave on a 450 miles trip in order to attend a meeting that will start 10.8 hours after you begin your trip. Along the way y

konstantin123 [22]

Answer:

c. 2.6 h

Explanation:

The longest time spent over dinner is the time that you have available minus the minimum possible time spent in the trip.

The time of the trip is found using:

t = $\frac{d}{v}$

Where distance is d and velocity is v. The time will be minimum at maximum velocity. Replacing with the data we have:

Ttrip = $\frac{450}{55}$ = 8.1818 h

Tdinner = 10.8h - 8.1818 h = 2.6181h

that aproximates 2.6 h.

4 0

3 years ago

The isomerization of cyclopropane to form propene is a first-order reaction. At 760 K, 85% of a sample of cyclopropane changes t

Nataliya [291]

Answer:

so rate constant is 4.00 x 10^-4 $s^{-1}$

Explanation:

Given data

first-order reactions

85% of a sample

changes to propene t = 79.0 min

to find out

rate constant

solution

we know that

first order reaction are

ln [A]/[A]0 = -kt

here [A]0 = 1 and (85%) = 0.85 has change to propene

so that [A] = 1 - 0.85 = 0.15.

that why

[A] / [A]0= 0.15 / 1

[A] / [A]0 = 0.15

here t = (79) × (60s/min) = 4740 s

k = - {ln[A]/[A]0} / t

k = -ln 0.15 / 4740

k = 4.00 x 10^-4 $s^{-1}$

so rate constant is 4.00 x 10^-4 $s^{-1}$

3 0

3 years ago

Which of the following is required to change the scientific knowledge base

Ksenya-84 [330]

Scientific knowledge is based upon observation, and it is supplemented by experimentation.<span> Scientific research follows the scientific method, a four-step process that guides scientists in the accumulation of knowledge.</span>

5 0

3 years ago

A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan

Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

$E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C$

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

$E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C$

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

6 0

3 years ago