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3 years ago

According to Newton's first law of motion, what will an object in motion do when

Physics

2 answers:

ivann1987 [24]3 years ago

6 0

It could speed up, change direction, come to a complete stop, or slow down without stopping.

The one thing the object could NOT do is keep moving with the same velocity. Its velocity must change.

kolezko [41]3 years ago

4 0

Answer:

The Answers B, move at the same velocity

Explanation:

bruh

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An ice cube can slide around the inside of a vertical circular hoop of radius . It undergoes small-amplitude oscillations if dis

klasskru [66]

Answer:

$T=2\pi \times \sqrt {\frac {g}{\mu R}$

Explanation:

$\mu mw^{2}R=mg$

where m is the mass, g is acceleration due to gravity

The masses m are on both sides hence they cancel

$W=\sqrt {\frac {g}{\mu R}$

We know that T=2\pi W and substituting W with the above equation then

$T=2\pi \times \sqrt {\frac {g}{\mu R}$

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4 years ago

Pulling a rubber band back and then letting it fly across the room is an example of

Cloud [144]

Answer:

Explanation:

because a elastic band uses elastic energy

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3 years ago

A circular loop of flexible iron wire has an initial circumference of 164cm , but its circumference is decreasing at a constant

Travka [436]

Answer:

emf = 0.02525 V

induced current with a counterclockwise direction

Explanation:

The emf is given by the following formula:

$emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}$ $\ \ =-B\frac{A_2-A_1}{t_2-t_1}$ (1)

ФB: magnetic flux = BA

B: magnitude of the magnetic field = 1.00T

A2: final area of the loop; A1: initial area

t2: final time, t1: initial time

You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.

In t = 4 s the final circumference will be:

$c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm$

To find the areas A1 and A2 you calculate the radius:

$r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm$

r1 = 0.261 m

r2 = 0.190 m

Then, the areas A1 and A2 are:

$A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2$

Finally, the emf induced, by using the equation (1), is:

$emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV$

The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.

4 0

3 years ago

A 1 in diameter solid round bar has a groove 0.1 in deep with a 0.1 in radius machined into it. The bar is made of AISI 1040 CD

Maru [420]

The answer & explanation for this question is given in the attachment below.

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4 years ago

In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom

krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2 (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R = 38.5 x ( v^2 )

77 x 9.8 x 3.65 = 38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) = 71.54

v = 8.4 m/s ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0

4 years ago