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Anastaziya [24]
3 years ago
15

Is this unbalanced force or is it balanced also is it technically acceleration?

Physics
1 answer:
Ivan3 years ago
7 0

Answer:

Unbalanced

Explanation:

There is an acceleration. Look at the direction of the arrows.

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A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F
Marrrta [24]

Answer:

Part a)

x = 0.4 m

Part b)

v_i = 11.7 m/s

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

\frac{1}{2}mv^2 = \frac{1}{2}kx^2

\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2

x = 0.4 m

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

W_f = -7( \pi R)

W_f = -7(\pi \times 1.00)

W_f = -22 J

work done against gravity is given as

W_g = - mg(2R)

W_g = -(0.500)(9.8)(2\times 1)

W_g = -9.8 J

Now by work energy equation we have

\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2

\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2

v_f = 4.1 m/s

Part c)

now minimum speed required at the top is such that the normal force must be zero

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = 3.13 m/s

so here we got speed more than the required speed so it will reach the top

5 0
3 years ago
Could an<br> average star, such as our<br> sun, become a neutron star?<br> Explain your answer.
larisa86 [58]
<span>No. Neutron stars are the remnants of very large stars that have supernova'd. Anything below 1.44 solar masses becomes a dwarf, anything above 5 solar masses becomes a black hole. Everything in between becomes a neutron star (or quark star, but it's not proven).</span>
5 0
3 years ago
A pilot heads his jet due east. The jet has a speed of 475 mi/h relative to the air. The wind is blowing due north with a speed
oksian1 [2.3K]

Explanation:

a. The velocity of the wind as a vector in component form will be represented as v vector:

    v=30j

b.The velocity of the jet relative to the air as a vector in component form will be represented as u vector

    u=475i

c. The true velocity of the jet as a vector will be represented as w:

  w=u+v

  w=475i+30j

d.  The true speed of the jet will be calculated as:

    IwI=\sqrt{(475)^2+(30)^2}

    IwI=\sqrt{225625+900}

    IwI=\sqrt{226525}

    IwI=476 mi/h

e. The direction of the jet will be:

tita=tan^{-1}\frac{30}{475}

tita=tan^{-1}(0.0632)

tita=3.62degrees,or,N86.38degreesS

7 0
3 years ago
A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a
Arlecino [84]

Answer:

<em><u>M</u></em><em><u>a</u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u>m</u></em><em><u>a</u></em><em><u>t</u></em><em><u>i</u></em><em><u>c</u></em><em><u>a</u></em><em><u>l</u></em><em><u>l</u></em><em><u>y</u></em><em><u>:</u></em>

That will be

<em>=</em><em> </em><em>1</em><em>5</em><em>0</em><em>0</em><em> </em><em>x</em><em> </em><em>1</em><em>5</em><em> </em><em>x</em><em> </em><em>4</em><em>5</em><em>0</em><em>0</em>

<em>=</em><em> </em><em><u>1</u></em><em><u>0</u></em><em><u>1</u></em><em><u>,</u></em><em><u>2</u></em><em><u>5</u></em><em><u>0</u></em><em><u>,</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

5 0
3 years ago
A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853
Norma-Jean [14]

Recall that

{v_f}^2-{v_i}^2=2a\Delta x

where v_i and v_f are the initial and final velocities, respecitvely; a is the acceleration; and \Delta x is the change in position.

So we have

\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)

\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

7 0
3 years ago
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