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Darya [45]
1 year ago
10

A disk is spinning about its center with a constant angular speed at first. Let the turntable spin faster and faster, with const

ant angular acceleration ?. Which sketch qualitatively shows the direction of the acceleration vector a of the bug?

Physics
1 answer:
hoa [83]1 year ago
7 0

Answer:

4 (please see the attached file)

Explanation:

While the angular speed (counterclockwise) remained constant, the angular acceleration was just zero.

So, the only force acting on the bug (parallel to the surface) was the centripetal force, producing a centripetal acceleration directed towards the center of the disk.

When the turntable started to spin faster and faster, this caused a change in the angular speed, represented by the appearance of an angular acceleration α.

This acceleration is related with the tangential acceleration, by this expression:

at = α*r

This acceleration, tangent to the disk (aiming in the same direction of the movement, which is counterclockwise, as showed in the pictures) adds vectorially with the centripetal force, giving a resultant like the one showed in the sketch Nº 4.

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Kayla and her friends are setting up chairs for a school play each row will contain the same number of chairs Kayla knows that t
LUCKY_DIMON [66]

Answer:

96=8*c

Explanation:

4 0
1 year ago
(a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the ang
avanturin [10]

6m at 0.9 radian means 6m at 51.57^0

Since positive angles are counter clockwise

6m at 51.57^0 can be written as 6*cos51.57^0i+6*sin51.57^0j = 3.73i+4.70j

5m at -75^0 can be written as 5*cos(-75)^0i+5*sin(-75)^0j = 1.294i-4.83j

4m at 1.2 radian means 4m at 68.75^0

4m at 68.75^0 can be written as 4*cos68.75^0i+4*sin68.75^0j = 1.45i+3.73j

6m at -210^0 can be written as 6*cos(-210)^0i+6*sin(-210)^0j = -5.20i-3j

a) So sum of the vector = 1.274i+0.6j

b) Magnitude = \sqrt{1.274^2+0.6^2} = 1.98 m

c) Angle,  tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0

7 0
1 year ago
A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h
Anestetic [448]

Answer:

Speed of 1.83 m/s and 6.83 m/s

Explanation:

From the principle of conservation of momentum

mv_o=m(v_1 + v_2) where m is the mass, v_o is the initial speed before impact, v_1 and v_2 are velocity of the impacting object after collision and velocity after impact of the originally constant object

5m=m(v_1 +v_2)

Therefore v_1+v_2=5

After collision, kinetic energy doubles hence

2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})

2v_o^{2}=v_1^{2} + v_2^{2}

Substituting 5 m/s for v_o then

2*(5^{2})= v_1^{2} + v_2^{2}

50= v_1^{2} + v_2^{2}

Also, it’s known that v_1+v_2=5 hence v_1=5-v_2

50=(5-v_2)^{2}+ v_2^{2}

50=25+v_2^{2}-10v_2+v_2^{2}

2v_2^{2}-10v_2-25=0

Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then v_2=6.83 m/s

Substituting, v_1=-1.83 m/s

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s

6 0
1 year ago
A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.
myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.  You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain  the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

5 0
1 year ago
A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th
son4ous [18]

Answer:

Part a)

E = 8436.7 N/C

Part b)

E = 8436.7 N/C

Explanation:

Part a)

Electric field due to large sheet is given as

E = \frac{\sigma}{2\epsilon_0}

\sigma = \frac{Q}{A}

Q = -3.8 nC

A = \pi(0.09)^2

A = 0.025 m^2

\sigma = \frac{-3.8\times 10^{-9}}{0.025}

\sigma = -1.5 \times 10^{-7} C/m^2

now the electric field is given as

E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}

E = 8436.7 N/C

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

E = 8436.7 N/C

5 0
2 years ago
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