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lilavasa [31]
2 years ago
5

A box of mass m is pulled with a constant acceleration a along a horizontal frictionless floor by a wire that makes an angle of

15° above the horizontal. If T is the tension in this wire, then _____________.
A) T = ma.
B) T > ma.
C) T < ma.
Physics
1 answer:
emmainna [20.7K]2 years ago
6 0

Answer:

B) T > ma.

Explanation:

To solve this problem, we have to analyze the forces acting in the horizontal direction.

In the horizontal direction, we have:

- The horizontal component of the tension in the wire, Tcos \theta, where T is the magnitude of the tension and \theta the angle that the wire makes with the horizontal

Since this is the only force acting on the box in the horizontal direction, this is also the net force, so it is equal to the product of mass and acceleration (Newton's second law of motion):

Tcos \theta = ma

where

m is the mass of the box

a is the acceleration

We can rewrite the equation as

T=\frac{ma}{cos \theta}

The angle in this problem is \theta=15^{\circ}, so

T=\frac{ma}{cos 15^{\circ}}=\frac{ma}{0.966}=1.035 ma

Therefore, the correct option is

B) T > ma.

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A person standing for a long time gets tired when he does not appear to do any work .why?​
Keith_Richards [23]

Explanation:

A person standing still for a long time feels tired because the force of gravity acts on our body and puts stress on our muscles. so our muscles need energy to do work and keep body balanced and help to stand upright.

6 0
1 year ago
If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal
iVinArrow [24]

Answer:

xcritical = d− m1 /m2 ( L /2−d)

Explanation: the precursor to this question will had been this

the precursor to the question can be found online.

ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)

. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces

smallest possible value of x such that the bar remains stable (call it xcritical)

∑τA = 0 = m2g(d− xcritical)− m1g( −d)

xcritical = d− m1 /m2 ( L /2−d)

6 0
1 year ago
A community calendar allows nonprofit organizations to add events and their intended purposes for community members to see.
lbvjy [14]

Out of all given choices, activities and donor list can be found on the given kind of informative website.

Answer: Option A and B

<u>Explanation:</u>

Non-profit need to do the examination and for sake activities that simply aren't successful. And afterwards, they have to look to a portion of these auxiliary changes that to discover increasingly productive approaches to make a feasible money related model for their social change work.

When occasions are crucial, allowed to visit, and concentrated on developing as well as managing present or potential significant donors (people, establishments, corporate pioneers, they can bode well. Yet, only in the event that you catch up with participants on a one-on-one premise to additionally put them in the association and in the long run request that they contribute or reestablish their commitments.

On the off chance, you don't charge those to visit with the goal that can approach them for a greater, and progressively important blessing not far off.

8 0
2 years ago
In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequ
grigory [225]

Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

I = \frac{1}{2}ε₀cE_{max²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E_{max is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E_{max  )

E_{max = 350/f kV/m

given that frequency of 60 Hz, we substitute

E_{max = 350/60 kV/m

E_{max = 5.83333 kV/m

E_{max = 5.83333 kV/m × ( \frac{1000 V/m}{1 kV/m} )

E_{max = 5833.33 N/C

so we substitute all our values into the formula for  intensity of an electromagnetic wave;

I = \frac{1}{2}ε₀cE_{max²

I = \frac{1}{2} × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

I = 45 × 10³ W/m²

I = 45 × 10³ W/m² × ( \frac{1 kW/m^2}{10^3W/m^2} )

I = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

7 0
1 year ago
carbon-14 has a half-life of approximately 5,700 years. a fossil shell contain 25% of the original amount of its carbon-14. appr
denis23 [38]

The half-life equation m=m_{0} (\frac{1}{2})^n in which <em>n </em>is equal to the number of half-lives that have passed can be altered to solve for <em>n.</em>

<em>n = \frac{log(\frac{m}{m_{0}} )}{log(\frac{1}{2})}</em>

<em>\frac{log(\frac{.25}{1} )}{log(\frac{1}{2})} = 2</em>

Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.

<em>2 * 5700 =  </em>11400 yr

3 0
2 years ago
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