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shusha [124]
4 years ago
8

Amy runs 2 blocks south, then turns around and runs 3 blocks north. what is the distance and displacement

Physics
1 answer:
Elenna [48]4 years ago
5 0

Answer: Distance=  5 blocks and displacement= -1 block

Explanation: distance is the total length of a journey while displacement is the shortest straight length of a journey!

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A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.260
ruslelena [56]

Answer:

The tension in the rope is 41.38 N.

Explanation:

Given that,

Mass of bucket of water = 14.0 kg

Diameter of cylinder = 0.260 m

Mass of cylinder = 12.1 kg

Distance = 10.7 m

Suppose we need to find that,

What is the tension in the rope while the bucket is falling

We need to calculate the acceleration

Using relation of torque

\tau=F\times r

I\times\alpha=F\times r

Where, I = moment of inertia

\alpha = angular acceleration

\dfrac{Mr^2}{2}\times\dfrac{a}{r}=F\times r

F=\dfrac{M}{2}a...(I)

Here, F = tension

The force is

F=m(g-a)...(II)

Where, F = tension

a = acceleration

From equation (I) and (II)

\dfrac{M}{2}a=m(g-a)

a=\dfrac{g}{1+\dfrac{M}{2m}}

Put the value into the formula

a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}

a=6.84\ m/s^2

We need to calculate the tension in the rope

Using equation (I)

F=\dfrac{M}{2}a

Put the value into the formula

F=\dfrac{12.1}{2}\times6.84

F=41.38\ N

Hence, The tension in the rope is 41.38 N.

6 0
3 years ago
Please help I’m so confused
Ilya [14]

<em>1.wavelength</em>

<em>2.trough</em>

<em>3.amplitude</em>

<em>4.crest</em>

5 0
3 years ago
A beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a
Verizon [17]

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

20

1

+

12

1

⟹

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

−16

1

+

12

1

⟹

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

6 0
3 years ago
Một xe máy đang đi với v = 50,4 km/h bỗng người lái xe thấy có ổ gà trước mắt cách xe 24,5 m.Người ấy phanh gấp và xe đến ổ gà t
Finger [1]

Answer:

<h3>эмне, кечиресиз айым, бирок мен сиздин тилиңизди түшүнбөй</h3>
3 0
3 years ago
If a box of supplies is dropped from the cargo hold of an airplane travelling at an altitude of 4,410 meters how long will it ta
Paraphin [41]

Answer:

It will take 30 seconds to reach the ground, and it will be travelling at 294 m/s when it does so. This means that its average velocity was 147 m/s.

Explanation:

d=v_ot+\dfrac{1}{2}at^2

Since the initial velocity of a dropped object is 0, we can make this the equation:

d=\dfrac{1}{2}at^2 \\\\4410=\dfac{1}{2}(9.8)t^2 \\\\t^2=900

t=30\text{ seconds}

The final velocity can be calculated with the formula:

v_f=v_o+at

Once again, since there is no initial velocity:

v_f=at \\\\v_f=(9.8)(30)=294m/s

Since the initial velocity is 0, the average vertical velocity is 294/2=147 m/s.

Hope this helps!

5 0
3 years ago
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