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3 years ago

A person tries to lift each object with a force of 25 N, upward. Which

Physics

2 answers:

Vlad1618 [11]3 years ago

4 0

Answer:

Option C. Objects 1 and 3 will not move, and objects 2 and 4 will accelerate

upward.

Explanation:

The following data were obtained from the question:

OBJECT >>>>>>>>> WEIGHT (N)

1 >>>>>>>>>>>>>>>> 35

2 >>>>>>>>>>>>>>>> 23

3 >>>>>>>>>>>>>>>> 26

4 >>>>>>>>>>>>>>>> 18

Force (F) applied = 25 N

From the above, the force applied to each object is 25N. Thus the following can be concluded based on the data given above:

For object 1:

Weight = 35 N

Force applied = 25 N

Thus, the object will not move since the weight of the object is greater than the force applied

For object 2:

Weight = 23 N

Force applied = 25 N

Thus, the object will move since the force applied is greater than the weight of the object.

For object 3:

Weight = 26 N

Force applied = 25 N

Thus, the object will not move since the weight of the object is greater than the force applied.

For object 4:

Weight = 18 N

Force applied = 25 N

Thus, the object will move since the force applied is greater than the weight of the object.

From the above illustrations, Object 1 and 3 will not move, and objects 2 and 4 will accelerate i.e move

inn [45]3 years ago

3 0

Answer:

Explanation:

Guy above was correct

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If A > B, under what condition is |A-BI=|A|- IB|? a. Vectors A and B are in opposite directions b. Vectors A and B are in the

kupik [55]

Answer:

b) Vectors A and B are in the same direction.

Explanation:

To understand this problem we will say that vector A has a magnitude of 5 units and vector B a magnitude of 3 units. In the subtraction of vectors the initial parts of vectors always bind together. And the vector resulting from the subtraction is traced from the end of the second vector (B) to the end of the first vector (A).

The length of the resultant vector will be 5 - 3 = 2

In the attached image, we analyze case a), b), and d)

For a)

As we can see in the attached image the resultant vector has a length of 8 units.

For d)

As we can see in the attached image the resultant vector has a length of 5.83 units.

For b)

The resultant vector has a length of 2 units.

Therefore the case given in b) is true

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3 years ago

Changing the length of an air column will alter its _______

bija089 [108]

Natural Frequency
Hope that helped!

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3 years ago

Read 2 more answers

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

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3 years ago

Which of the following occurs with both a cold front and a mountain breeze?

BigorU [14]

Answer:

b warm air rises

Explanation:

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3 years ago

What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0.30?

lorasvet [3.4K]

The momentum of the x-ray photon is p = h/lambda . Lambda is the wavelength (0.30nm=3x10^(-9)m) and h is Planck's constant,(h=6.62607004 × 10-34<span> m2 kg / s).The momentum is: 2.2 x 10^(-25).</span>

The momentum can be calculated also as: p=mv, where m is the mass of the electron and v is the speed.

So v=p/m,p is known,and also the mass of the electron (m=9.10938356 × 10-31<span> kilograms).</span>

v=2.2 x 10^(-25)/9.10938356 × 10-31<span> kilograms=0.24 x 10^6 m/s</span>

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3 years ago