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Vedmedyk [2.9K]
3 years ago
5

A force of 120 N is applied to the front of a sledge at an angle of 28.00 above the horizontal so as to pull the sledge a distan

ce of 165 meters. How much work was done by the applied force?
Physics
1 answer:
PtichkaEL [24]3 years ago
4 0

Answer:

Workdone = 17482.36 Joules

Explanation:

Given the following data;

Force, F = 120 N

Angle, d = 28.0°

Distance, x = 165 m

To find the work done, we would use the following formula;

Workdone = FxCosd

Substituting into the formula, we have;

Workdone = 120*165*Cos(28)

Workdone = 19800 * 0.8830

Workdone = 17482.36 Joules

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mel-nik [20]

Answer:

\mathbf{f_o =1.87 \times 10^{11} \ Hz}

Explanation:

From the given information:

The elastic potential energy can be calculated by using the formula:

U_o = \dfrac{1}{2}kR_o^2

Making K the subject;

K = \dfrac{2 U_o}{R_o^2}

k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}

k = 2.3 × 10⁻² N/m

Now; the frequency of the small oscillation can be determined by using the formula:

f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}

where;

m = mass of each atom = 1.66 × 10⁻²⁶ kg

f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}

\mathbf{f_o =1.87 \times 10^{11} \ Hz}

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3 years ago
How did the invention of the printing press affect society
Ilia_Sergeevich [38]
The printing press had dramatic effects on European civilization. Its immediate effect was that it spread information quickly and accurately. This helped create a wider literate reading public.
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3 years ago
An acetal polymer block is fixed to the rigid plates at its top and bottom surfaces. If the top plate displaces 2 mm horizontall
Taya2010 [7]

Answer:

The shear modulus of the polymer is 5 MPa

Solution:

As per the question:

displacement of the top plate, d = 2 mm = 0.002 m

Horizontal Force, P = 2 kN = 2000 N

Width of the block, w = 100 mm = 0.1 m

By small angle analysis:

Angle, \gamma = \frac{2}{200} = 0.01 rad

Now,

Stress, \tau = \frac{P}{Area} = \frac{2000}{0.4\times 0.1} = 50000 Pa

G = \frac{\tau}{\gamma} = \frac{50000}{0.01} = 5 MPa

3 0
3 years ago
The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a
sertanlavr [38]

Answer:

Electric field, E = 6.36\times 10^{6}\ V/m

Explanation:

Given that,

The potential difference across the membrane, \Delta V=0.063\ V

The thickness of the membrane is, \Delta x=9.9\times 10^{-9}\ m

We need to find the magnitude of the electric field in the membrane. We know that the relation between electric field and electric potential is given by :

E=\dfrac{-\Delta V}{\Delta x}\\\\E=\dfrac{0.063\ V}{9.9\times 10^{-9}\ m}\\\\E=6.36\times 10^{6}\ V/m

So, the magnitude of the electric field in the membrane is 6.36\times 10^{6}\ V/m.

7 0
4 years ago
You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testt
sertanlavr [38]

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

f_{1}=\dfrac{v}{4l}

Put the value into the formula

f_{1}=\dfrac{344}{4\times0.11}

f_{1}=781.81\ Hz

f_{1}=0.782\ kHz

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

f_{1}=\dfrac{v}{4(\dfrac{L}{2})}

Put the value into the formula

f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}

f_{1}=1563.63\ Hz

f_{1}=1.563\ kHz

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

6 0
4 years ago
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