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HACTEHA [7]
3 years ago
11

A toolbox of mass 3.2kg is lowered by a rope from the roof to the ground. Find the acceleration of the toolbox when the force of

the rope is: a) 40.0 N b) 20.0N
answer: a) 2.7m/s^2, up b) 3.6m/s^2, down

I would like an explanation or equation to get to that answer. Thank you!
Physics
1 answer:
Lisa [10]3 years ago
8 0

Answer:

The answer to your question is:

a) 2.7 m/s²

b) -3.6 m/s²

Explanation:

Data

mass of the toolbox = 3.2 kg

a = ?

F = 40 N and F = 20 N

g = 9.81 m/s²

Formula

Second law of motion = F = ma

                              a + g = F / m

                              a = F/m - g

a)                            a = 40/3.2 - 9.81

                              a = 2.69 ≈ 2.7 m/s²   positive up

b)                            a = 20/ 3.2 - 9.81

                              a = 6.25 - 9.81

                                  = - 3.56 ≈ - 3.6 m/s²  negative down

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2 years ago
Which variable mentioned in Table 2-1 is kept constant? a. amount of time spent swimming b. type of swimming stroke c. number of
Vladimir [108]

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a. amount of time spent swimming

7 0
2 years ago
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
If a metal wire is 4m long and a force of 5000n causes it to stretch by 1mm, what is the strain?
barxatty [35]

Answer:

2.5\cdot 10^{-4}

Explanation:

The strain is defined as the ratio of change of dimension of an object under a force:

S=\frac{\Delta L}{L_0}

where

\Delta L is the change in length of the object

L_0 is the original length of the object

In this problem, we have L_0 = 4 m and \Delta L=1 mm=0.001 m, therefore the strain is

S=\frac{\Delta L}{L_0}=\frac{0.001 m}{4 m}=2.5\cdot 10^{-4}


5 0
3 years ago
His eyes are 1.76 m above the floor, and the top of his head is 0.15 m higher. Find the height (in m) above the floor of the top
krek1111 [17]

Answer:

The height (in m) above the floor of the top and bottom of the smallest mirror in which he can see both the top of his head and his feet is;

1.835 m and 0.88 m.

Explanation:

Here we have the total height of the man as

1.76 + 0.15 = 1.91 m

The mirror is positioned such that the person can see both the top of his head and his feet

We have the eyes are 0.15 m below the top of the head, therefore by the law of reflection, the incident and reflected angle must be equal.

Hence, the light from the top of his head and then reflected to his eyes forms a isosceles triangle, with the base being the distance of the eye to the top of his head and the top of the triangle is on the mirror.

The height of the mirror is then

1.91 - 0.15/2 = 1.835 m

Similarly, the distance from the eye to the feet is 1.76, therefore, the base of the mirror is positioned at 1.76/2 or 0.88 m above the ground.

7 0
3 years ago
Read 2 more answers
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