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zheka24 [161]
3 years ago
12

A train is traveling down a straight track at 20. m/s when the engineer applies the brakes, resulting in an acceleration of -1.0

m/s2 as long as the train is in motion. how far does the train move during a 40. s time interval starting at the instant the brakes are applied? [double-check that your answer makes logicalsense.
Physics
1 answer:
Kipish [7]3 years ago
7 0

Answer:

The answer to your question is: d = 0 m, it does not move

Explanation:

Data

vo = 20 m/s

a = -1 m/s2

t = 40 s

d = ?

Formula

d = vot + (1/2)at²

Substitution

d = (20)(40) + (1/2)(-1)(40)²

d = 800 - 800

d = 0 m                                  It suggest that it does not move.

                                       I hope it can help you

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⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)
GREYUIT [131]

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

7 0
3 years ago
Teams A and B are in a tug-of-war challenge. Team A wins the challenge. What can be said about Team A?
Harrizon [31]
Team a was physically stronger?
7 0
3 years ago
Read 2 more answers
What acceleration will you give to a 24.5 kg
Ludmilka [50]

Heya!!

For calculate aceleration, lets applicate second law of Newton:

                                                   \boxed{F=ma}

                                                 <u>Δ   Being   Δ</u>

                                             F = Force = 78,3 N

                                            m = Mass = 24,5 kg

                                             a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

\boxed{a=78,3\ N / 24,5\ kg}

⇒ Resolving

\boxed{a=3.19\ m/s^{2}}

Result:

The aceleration is <u>3,19 meters per second squared (m/s²)</u>

Good Luck!!

4 0
3 years ago
The ionosphere lies between the mesosphere and exosphere is it true or false
Helga [31]

False because the Ionsophere lies between the Mesosphere and the Theromsphere. If can can you give me brainliest :o ?

6 0
3 years ago
Which is not an example of a scalar?<br><br> a. 2t/s<br> b. 3kg<br> c. 6.2m north<br> d. -100c
xxMikexx [17]
C because I’m a teacher
4 0
3 years ago
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