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2 years ago

15

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to

an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly move at a constant velocity of 256m/s

1 answer:

Salsk061 [2.6K]2 years ago

7 0

Explanation:

$1.2 \mathrm{N} ; 2 \mathrm{N}$

2. $200 \mathrm{N} ; 200 \mathrm{N}$

4. $2 \mathrm{N} ; 2 \mathrm{N} ; 4 \mathrm{N}$

$5.2 \mathrm{N} ; 2 \mathrm{N} ; 2 \mathrm{N}$

$6.2 \mathrm{N} ; 2 \mathrm{N} ; 3 \mathrm{N}$

$8.200 \mathrm{N} ; 200 \mathrm{N} ; 5 \mathrm{N}$

In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of $256 \mathrm{m} / \mathrm{s}$

You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).

In both cases, the net force applied to the object must equal zero.

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A buoy is a floating device that can have many purposes, but often as a locator for ships. Collin constructs a hollow metal buoy

Yuri [45]

Answer:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water

B )angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

c ) h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

Explanation:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water this is because the substances with lesser density floats when placed in a substance with a higher density

B ) when the Buoy is at rest ( t = 0 ) and is then pushed down calculate angular frequency ( w ) of small oscillations in terms of the given variables

volume of cone = hA /3.

h = height of cone, A = Base Area.

therefore the total volume of the Buoy above water level ( at rest )

= ( $\pi r^{2} * \frac{h}{3}$ ) * 2 = $\frac{\pi r^{2} h }{6}$

The part of the buoy immersed in water = x

then the net upward force will be

fb ( force of Buoy ) = fg ( force of gravity )

note force = Mass * Acceleration

force of buoy = $\frac{\alpha }{2} *$ $\frac{\pi r^{2} }{h}$ * a = ( force of gravity ) $\alpha * T\frac{r^{2} }{4} * x * g$

therefore a = $\frac{3g}{h} * x$

angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

C ) height of the each cone

$\frac{2\pi }{w} = 1$ therefore w = 2 $\pi$

back to the angular frequency : $\frac{3g}{h} = 4\pi ^{2}$

therefore h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

4 0

2 years ago

A car travels due east 14 miles in 18 minutes. Then, the car turns around and returns to its starting point, taking an additiona

N76 [4]

Answer:

(a) The average speed is 0.85 milles/minute

(b) The average velocity is zero

Explanation:

In order to answer part (a) and (b) you have to apply the formulas for average speed and average velocity which are:

<em>-Average speed formula:</em>

$v=\frac{d}{t}$

where d is the total distance traveled and t is the tota time

Replacing the given values:

$v=\frac{14+14}{18+15}\\v=\frac{28}{33} \\v=0.85$ milles/minute

Notice that you have to replace the total distance, which is 14 milles for the go plus 14 milles for the return. The same for the total time.

<em>-Average velocity formula:</em>

V = Δx/Δt

Where V is the velocity vector, Δx is the displacement and Δt is the change in time

V= $\frac{X2-X1}{t2-t1}$

Where X2 is the final position and X1 is the initial position

In this case X1= 0 i and X2=0 i (i is the unit vector in the x direction). So, the displacement is zero.

Therefore, the average velocity is:

V= 0 i [milles/minute]

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2 years ago

The electric field strength E is measured as:

soldi70 [24.7K]

The correct answer is
<span>force per unit charge.

In fact, the electric field strength is defined as the electric force per unit charge experienced by a positive test charge located in the electric field. In formula:
</span> $E= \frac{F}{q}$
where
E is the electric field strength
F is the electric force experienced by the charge
q is the positive test charge.

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2 years ago

Everyone open this please!

notsponge [240]

Answer:

yeah i knowwwwwwwwwwwwww

Explanation:

7 0

2 years ago

Read 2 more answers

How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n

grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

$dSin\theta = m\lambda$

Where

m= Any integer which represent the number of repetition of spectrum

$\lambda$ = Wavelength

d = Distance between the slits.

$\theta$ = Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

$d = \frac{m\lambda}{sin\theta }$

$d = \frac{2*536*10^{-9}}{sin(24)}$

$d = 2.63*10^{-6}m$

Therefore the number of lines per millimeter would be given as

$\frac{1}{d} = \frac{1}{2.63*10^{-6} }$

$\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})$

$\frac{1}{d} = 379.4 lines/mm$

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

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2 years ago