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FrozenT [24]
4 years ago
9

You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3

,500 N across the carpet to a spot 5 m away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dresser?
W> 1.75 x 10^4 J
W = 1.75 x 10^4 J
1.75 x 10^4 J > W > 0 J
W = 0 J
Physics
1 answer:
Mazyrski [523]4 years ago
8 0

Answer:

0.

Explanation:

(A) Work done on the dresser which will be given as :

W = F d cos \theta

where, F = weight of the dresser = 3500 N

d_{expected} = 5m

However you can move the dresser, that is a real distance,

d = 0 m

Substituting,

W = (0) F(3500)cos (\theta) (\theta is the angle at which you apply the force)

W = 0 J

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What does every magnet possess?
IRINA_888 [86]

Answer:

a North and South Pole :)

Explanation:

5 0
4 years ago
Anxious to the point that he couldn't eat very well without his stomach feeling queasy. As a result of his stress, Jonah increas
Mice21 [21]

Based on this information, a psychologist would most likely conclude that Jonah's stress led him toward positive personal changes.

<h3>What is stress?</h3>

When a person is afraid or tensed about something, he overthinks about him and become stressed.

Anxious to the point that he couldn't eat very well without his stomach feeling queasy. As a result of his stress, Jonah increased his time of practicing basketball and asked an old team member to help him develop new skills he needed in order to make the team. After hearing that he made the basketball team a day after tryouts, Jonah went out for a large pizza with the rest of the team players.

As, he is feeling good, he wants to celebrate himself. So, he went to eat pizza with his team players. This shows that he is not stressed but happy.

Thus, ,this information depicts that Jonah's stress led him toward positive personal changes.

Learn more about stress.

brainly.com/question/9643296

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6 0
2 years ago
(TCO 5) The relationship between Celsius (º C) and Fahrenheit (º F) degree of measuring temperature is linear. Find the linear e
gtnhenbr [62]

Answer:

C=-\dfrac{10}{17}(F-32)

Explanation:

Given that

32° F corresponds to 0 °C.  ---Point 1

212° F corresponds to 100 °C.----Point 2

We know that if two point is given that equation of line can be found as

y-y_1=\dfrac{y_2-y_1}{x_2-x_2}(x-x_1)

Lets C in y- direction and F in x- direction,so we can say that

C-C_1=\dfrac{C_2-C_1}{F_2-F_2}(F-F_1)

C-0=\dfrac{100-0}{32-212}(F-32)

C=-\dfrac{10}{17}(F-32)

So the linear relationship is

C=-\dfrac{10}{17}(F-32)

8 0
3 years ago
A polonium isotope with an atomic mass of 211.988868 u undergoes alpha decay, resulting in a daughter isotope with an atomic mas
prisoha [69]

Answer:

 K = 9.53 MeV

Explanation:

The kinetic energy that the alpha particle has emitted, is the energy in excess after removing the resting energy of the atoms and the helium nucleus that forms the alpha particle

Since energy and masses are related and cannot be

          m₀ c² = m_{f} c² + m_He c²+ K

          K = c² (m₀ - m_{f} - m_He)

           

the mass of the Helium atom is 4 u

           K = (3 10⁸)² (211,988868 -207.976652 - 4,002) 1,661 10⁻²⁷

           K = 14,949 10⁻¹¹ (0.0102)

             

          K = 1,527 10⁻¹² J

let's reduce 1 J = 6,242 10¹² MeV

           K = 9.53 MeV

7 0
4 years ago
A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o
Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

\delta = \frac{PL}{AE}

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

P = 68*10^3 N

E = 200GPa  

A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2

Replacing we have,

\delta = \frac{PL}{AE}

\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}

\delta = 0.001932m

\delta = 1.93mm

Therefore the change in length is 1.93mm

7 0
3 years ago
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