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mrs_skeptik [129]
3 years ago
9

A woman climbs up a ladder in 1.37 s at 2.20 m/s. How tall is the ladder?

Physics
2 answers:
kaheart [24]3 years ago
7 0

Answer:

height = 3.014

Explanation:

to find the height, use the equation height = velocity * time

height = 2.2*1.37

height =3.014

ArbitrLikvidat [17]3 years ago
6 0

Answer:

The ladder is 3.014 m tall.

Explanation:

To solve this problem, we must use the following formula:

v = x/t

where v represents the woman’s velocity, x represents the distance she climbed (the height of the ladder), and t represents the time it took her to move this distance

If we plug in the values we are given for the problem, we get:

v = x/t

2.20 = x/1.37

To solve this equation for x (the height of the ladder), we must multiply both sides by 1.37. If we do this, we get:

x = (2.20 * 1.37)

x = 3.014 m

Therefore, the ladder is 3.014 m tall.

Hope this helps!

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A box falls off of a tailgate and slides along the street for a distance of 62.5 m. Friction slows the box at –5.0 m/s2. At what
Mila [183]

Answer:

25 m/s

Explanation:

This question can be solved using equation of motion

v^2 = u^2 + 2as

where

v is the final velocity

u is the initial velocity

s is the distance covered while moving from initial to final velocity

a is the acceleration

_____________________________________________

Given

box moved for distance of 62.5 m

Friction slows the box at –5.0 m/s2----> this statement means that there is deceleration , speed of truck decreases by 5 m/s in every second until the box comes to rest. Friction causes this deceleration.

thus in this problem

a = -5.0 m/s2

V = 0   as body came to rest due to friction deceleration

u the initial velocity we have to find

the initial velocity of box will be the same as speed of truck, as the box was in the truck and hence box will pick the speed of truck.

so if we find speed of box, we will be able get sped of truck as well.

using equation of motion

v^2 = u^2 + 2as\\0^2 = u^2 + 2*-5* 62.5\\0 = u^2 - 625\\u^2 = 625\\\sqrt{u^2} = \sqrt{625} \\u = 25

Thus, initial speed with the truck was travelling was 25 m/s.

3 0
3 years ago
A disc with a mass of 1 kg moves horizontally to the right with a speed of 7 m/s on a table with negligible friction when it col
Elodia [21]

Answer:

1.6 m/s

Explanation:

First you need to find the momentums of each disc by multiplying their velocities with mass.

disc 1: 7*1= 7 kg m/s

disc 2: 1*9= 9 kg m/s

Second, you need to find the total momentum of the system by adding the momentums of each sphere.

9+7= 16 kg m/s

Because momentum is conserved, this is equal to the momentum of the composite body.

Finally, to find the composite body's velocity, divide its total momentum by its mass. This is because mass*velocity=momentum

16/10=1.6

The velocity of the composite body is 1.6 m/s.

7 0
2 years ago
If a girl is standing in front of a smooth surface from which a sound is reflected, the girl may hear
gavmur [86]
Near the surface of reflection, reflected wave may interfere with incident wave leading to production of constructive and as well as destructive interference. This in turn, can result to resonance as well as enhancement of the sound intensity as the waves of reflection adds to incident wave. Therefore, the girl would higher intensity of reflected waves as compared to incident waves.

Therefore, statement A is correct.
3 0
3 years ago
Read 2 more answers
An object is dropped from rest from a 70.6 m tower. Air resistance is negligible. After 0.32 seconds, what is magnitude and dire
dem82 [27]

Answer:

<em>1,378.9ms²</em>

Explanation:

Given the following

Distance S = 70.6m

Time t = 0.32secs

Initial velocity = 0m/s

Required

Acceleration

Using the equation of motion

S = ut+1/2at²

Substitute

70.6 = 0+1/2a(0.32)²

70.6 = 0.0512a

a = 70.6/0.0512

a = 1,378.9

<em>Hence the acceleration is 1,378.9ms²</em>

7 0
2 years ago
A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie
Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, V_{Hall} = 20\ mV = 0.02\ V

Magnetic Field, B = 0.10 T

Hall emf, V'_{Hall} = 69\ mV = 0.069\ V

Now,

Drift velocity, v_{d} = \frac{V_{Hall}}{B}

v_{d} = \frac{0.02}{0.10} = 0.2\ m/s

Now, the expression for the electric field is given by:

E_{Hall} = Bv_{d}sin\theta                            (1)

And

E_{Hall} = V_{Hall}d

Thus eqn (1) becomes

V_{Hall}d = dBv_{d}sin\theta

where

d = distance

B = \frac{V_{Hall}}{v_{d}sin\theta}                      (2)

(a) When \theta = 90^{\circ}

B = \frac{0.069}{0.2\times sin90} = 0.345\ T

(b) When \theta = 60^{\circ}

B = \frac{0.069}{0.2\times sin60} = 0.398\ T

5 0
3 years ago
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