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3 years ago

10

a spring of original length 10 cm, Stretches to 12 cm, when a Force of 40 N is applied to It. what is the extension of the sprin

g when a Force of 26 N, is applied?

1 answer:

Mila [183]3 years ago

8 0

Answer:

2.6

Explanation:

Since the formula is k=F/X when you criscross it becomes KX=F,10X=26,then over ten which equals 2.6.

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After the driver first notices the obstacle, the car moves uniformly for a time interval t1−t0=t before the brakes are applied.

loris [4]

Answer:

V(t1-t0)

Explanation:

Moving 'uniformly' means constant velocity (speed). the formula for constant speed motion is $V = \frac{distance}{time}$ =( change in position/ change in time)

where,

V is speed

given in the statement :

change in time = t = t1-t0

let the constant speed be ' V '

disance = X = X1-X0

applying the above mentioned formula: V = $\frac{X}{t}$

V = X/t

X = Vt

the distance X1-X0 = Vt =V(t1-t0)

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3 years ago

Read 2 more answers

A "moving sidewalk" in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5

pishuonlain [190]

Answer:

a.14 s

b.70 s

Explanation:

a.Let the sidewalk moving in positive x- direction.

Speed of sidewalk relative to ground= $v_s=1m/s$

Speed of women relative to sidewalk=v=1.5m/s

The speed of women relative to the ground

$v_w=v_s+v=1+1.5=2.5m/s$

Distance=35 m

Time= $\frac{distance}{speed}$

Using the formula

Time taken by women to reach the opposite end if she walks in the same direction the sidewalk is moving= $\frac{35}{v_w}=\frac{35}{2.5}=14s$

b.If she gets on at the end opposite the end in part (a)

Then, we take displacement negative.

Speed of sidewalk relative to ground= $v_s=1m/s$

Speed of women relative to sidewalk=v=-1.5 m/s

The speed of women relative to the ground= $v_w=v_s+v=1-1.5=-0.5m/s$

Time= $\frac{-35}{-0.5}=70 s$

Hence, the women takes 70 s to reach the opposite end if she walks in the opposite direction the sidewalk is moving.

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3 years ago

A 1.3-kg model airplane flies in a circular path on the end of a 23-m line. The plane makes

storchak [24]

(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in

(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s

(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:

(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s

(c) The plane accelerates toward the center of the path with magnitude

<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²

(d) By Newton's second law, the tension in the line is

<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N

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3 years ago

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frozen [14]

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