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ANEK [815]
3 years ago
11

In 1990 Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much wo

rk did Arfeuille do on the object? (b) What magnitude force did he exert on the object during the lift, assuming the force was constant?
Physics
1 answer:
Crank3 years ago
7 0

Answer:

(a). The work done is 472 J.

(b). The force exerted is 2.76 kN.

Explanation:

Given that,

Distance = 17.1 cm

Mass of object = 281.5 kg

(a). We need to calculate the work done

Using formula of work done

W=F_{net}\dotc d

W=mg\cdot d

Where, m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

W=281.5\times9.8\times17.1\times10^{-2}

W=472\ J

The work done is 472 J.

(b). We need to calculate the force

Using action reaction principle

F=F'

F=mg

Put the value into the formula

F=281.5\times9.8

F=2.76\ kN

The force exerted is 2.76 kN.

Hence, (a). The work done is 472 J.

(b). The force exerted is 2.76 kN.

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Nina [5.8K]

Answer: 0.05\ mm

Explanation:

Given

Cross-sectional area of wire A_1=4\ mm^2

Extension of wire \delta l=0.1\ mm

Extension in a wire is given by

\Rightarrow \delta l=\dfrac{FL}{AE}

where, E=\text{Youngs modulus}

\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)

for same force, length and material

\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)

Divide (i) and (ii)

\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm

5 0
2 years ago
An experiment is performed to test the effect of three different types of water on the growth of a plant. The test is done by us
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I think the correct answer would be D. The tap water in the experiment is one the three test conditions of the independent variable, the type of water. The independent variable in a experiment is the one being manipulated or the one being changed. In this case, it is the type of water.
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3 years ago
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A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continue
enot [183]

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

7 0
3 years ago
A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency
Mice21 [21]

Answer:

T = 712.9 N

Explanation:

First, we will find the speed of the wave:

v = fλ

where,

v = speed of the wave = ?

f = frequency = 890 Hz

λ = wavelength = 0.1 m

Therefore,

v = (890 Hz)(0.1 m)

v = 89 m/s

Now, we will find the linear mass density of the wire:

\mu = \frac{m}{L}

where,

μ = linear mass density of wie = ?

m = mass of wire = 90 g = 0.09 kg

L = length of wire = 1 m

Therefore,

\mu = \frac{0.09\ kg}{1\ m}

μ = 0.09 kg/m

Now, the tension in wire (T) will be:

T = μv² = (0.09 kg/m)(89 m/s)²

<u>T = 712.9 N</u>

7 0
3 years ago
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