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2 years ago

A student walks 1.0 mi west and then 1.0 mi north. afterward, how far is she from her starting point?

1 answer:

7 0

1.0 mi because as he has covered 1 the disrance of 1 m and then 1 m in north so he can go straight 1m north so the distance is actually 1m from the starting point

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How long will it take a plane to fly 1256km<br> if it travels 500km/hr?

expeople1 [14]

Answer:

<h3>The answer is 2.51 s</h3>

Explanation:

The time taken can be found by using the formula

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

d is the distance

v is the velocity

From the question we have

$t = \frac{1256}{500} \\ = 2.512$

We have the final answer as

Hope this helps you

4 0

2 years ago

Which of the following is an example of the conclusion phase of the scientific method?

Burka [1]

Answer:

a scientist examines the results and answers the lab question- last choice

5 0

2 years ago

A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h

timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

$F_n = \frac{mv^2}{r}$

$N+mgcos\theta = \frac{mv^2}{r}$

Here,

Normal reaction of the ring is N and velocity of the ring is v

$N+mgcos\theta = \frac{mv^2}{r}$

$N+Wcos\theta = \frac{W}{g} (\frac{v^2}{r})$

$N+2cos30\° = \frac{2}{32.2}*\frac{10^2}{2}$

$N = 1.374lb$

PART B) Acceleration

$F_t = ma_t$

$-mgsin\theta = ma_t$

$-W sin\theta = \frac{W}{g} a_t$

$-2Sin30\° = (\frac{2}{32.2})a_t$

$a_T = -16.10ft/s^2$

Negative symbol indicates deceleration.

<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>

8 0

3 years ago

Match the correct sentence together.

Brut [27]

I’m pretty sure you times them so 1 with A, 2 with e, 3 with C, and 4 with B

6 0

1 year ago

Read 2 more answers

A wagon with an initial velocity of 2 m/s and a mass of 60 kg, gets a push with 150 joules of

Aleks04 [339]

Answer:

v_f = 3 m/s

Explanation:

From work energy theorem;

W = K_f - K_i

Where;

K_f is final kinetic energy

K_i is initial kinetic energy

W is work done

K_f = ½mv_f²

K_i = ½mv_i²

Where v_f and v_i are final and initial velocities respectively

Thus;

W = ½mv_f² - ½mv_i²

We are given;

W = 150 J

m = 60 kg

v_i = 2 m/s

Thus;

150 = ½×60(v_f² - 2²)

150 = 30(v_f² - 4)

(v_f² - 4) = 150/30

(v_f² - 4) = 5

v_f² = 5 + 4

v_f² = 9

v_f = √9

v_f = 3 m/s

7 0

2 years ago