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3 years ago

15

A 150-kg object takes 1.5 minutes to travel a 2,500-meter straight path. It begins the trip traveling 120 m/s and decelerates to

a velocity of 20 m/s. What was its acceleration?

2 answers:

podryga [215]3 years ago

7 0

I believe it is -1.11 m/s^2. I will let you know if its correct

Rom4ik [11]3 years ago

5 0

Your answer is -1.11m/s2. i know this cause i had it right on a test. hope this helps everyone else looking for the answer

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Choose the scientific notation that best represents the standard notation given. 6,840,000,000 m 68.4 x 108 m 6.84 x 108 m 6.84

faust18 [17]

ANSWER

$6.84\times 10^9$

EXPLANATION.

To write in scientific notation means , we write in form,

$k\times 10^n$

Where n is an integer and

$1\le k\:$

The given number is,

$6,840,000,000$

We move the decimal point backward to obtain our k, the n here is 9, because we moved backward 9 times.

This implies that,

$6,840,000,000=6.84\times 10^9$

The correct answer is D

8 0

3 years ago

Read 2 more answers

A car that weighs 1.0 x 10^4 N is initially moving at a speed of 38 km/h when the brakes are applied and the car is brought to a

hram777 [196]

Answer:

Part a) Force on car = 2833.84 Newtons

Part b) Time to stop the car = 3.8 seconds

Part c) Factor for stopping distance is 4.

Part d) Factor for stopping time is 1.

Explanation:

The deceleration produced when the car is brought to rest in 20 meters can be found by third equation of kinematics as

$v^2=u^2+2as$

where

v = final speed of the car ( = 0 in our case since the car stops)

u = initial speed of the car = 38 km/hr = $\frac{38\times 1000}{3600}=10.56m/s$

a = deceleration produces

s = distance in which the car stops

Applying the given values we get

$0^2=10.56^2+2\times a\times 20\\\\a=\frac{0-10.56^2}{2\times 20}\\\\\therefore a=-2.78m/s^2$

Now the force can be obtained using newton's second law as

$Force=\frac{Weight}{g}\times a$

Applying values we get

$Force=\frac{1.0\times 10^4}{9.81}\times -2.78\\\\\therefore F=-2833.84Newtons$

The negative direction indicates that the force is opposite to the motion of the object.

Part b)

The time required to stop the car can be found using the first equation of kinematics as

$v=u+at$ with symbols having the same meanings

Applying values we get

$0=10.56-2.78\times t\\\\\therefore t=\frac{10.56}{2.78}=3.8seconds$

Part c)

From the developed relation of stopping distance we can see that the for same force( Same acceleration) the stopping distance is proportional to the square of the initial speed thus doubling the initial speed increases the stopping distance 4 times.

Part d)

From the relation of stopping time and the initial speed we can see that the stopping distance is proportional initial speed thus if we double the initial speed the stopping time also doubles.

8 0

3 years ago

PLEASE HELP!!!!

Zina [86]

D is the answer
Imagine the magnetic field, it emanates from both

4 0

3 years ago

Explain what happens to the pitch of a cell phone ring when the amplitude of a sound wave increases. Justify your reasoning usin

dalvyx [7]

<span>Nothing happens to the pitch of a cell phone ring when the amplitude
of a sound wave increases.

Pitch and amplitude are both characteristics of a wave, but they're not
connected, and they don't influence each other.</span>

7 0

3 years ago

You are holding one end of an elastic cord that is fastened to a wall 3.0 m away. You begin shaking the end of the cord at 2.3 H

Karo-lina-s [1.5K]

Answer:

Time take to fill the standing wave to the entire length of the string is 1.3 sec.

Explanation:

Given :

The length of the one end $x=$ $3m$ , frequency of the wave $f$ = 2.3 Hz, wavelength of the wave λ = 1 m.

Standing wave is the example of the transverse wave, standing wave doesn't transfer energy in a medium.

We know,

∴ $v = f$ λ

Where $v =$ speed of the standing wave.

also, ∴ $v=\frac{x}{t}$

where $t =$ time take to fill entire length of the string.

Compare above both equation,

⇒ $t =$ $\frac{3}{2.3}$ sec

$t = 1.3sec$

Therefore, the time taken to fill entire length 0f the string is 1.3 sec.

7 0

3 years ago