Can some one tell the answers

a spring has a force constant of 100 n/m and an unstretched length of 0.07 m. one end is attached to a post that is free to rota

Digiron [165]

F equals 3N with respect to the circle's center, moving in the same direction as the centripetal acceleration.

<h3>How much centripetal force is there in a centrifuge?</h3>

Centripetal force is the force that pushes an item in the direction of its center of curvature. It is fundamental to how a centrifuge operates.

<h3>On a roller coaster, what is centripetal force?</h3>

An item travelling in a circle is pushed inward toward what is known as the center of rotation, which is essentially what a roller coaster accomplishes when it travels through a loop. The force that maintains an object moving along a curved route is this pull toward the center, or centripetal force.

To know more about centripetal force visit:-

brainly.com/question/11324711

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6 0

1 year ago

What is a newton equal to in terms of units of mass and acceleration

artcher [175]

1 newton is the force needed to accelerate 1 kilogram of mass
at the rate of 1 meter per second² .

1 N = 1 kg-m/s² .

It's a force equal to roughly 3.6 ounces.

7 0

3 years ago

Read 2 more answers

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago

What would the velocity of an object be if it was on a 1.75m

JulijaS [17]

Answer:

T = 4 sec / 2 = 2 sec period of revolution

S = 2 pi R = 2 * pi * 1.75 m = 11 m

V = S / T = 11 m / 2 sec = 5.5 m/s speed of object

4 0

2 years ago

1. What distance is required for a train to stop if its initial velocity is 23 m/s and its

Irina-Kira [14]

Answer:

x=?

dt=?

vi=23m/s

vf=0m/s (it stops)

d=0.25m/s^2

time =

vf=vi+d: 0=23m/s+(0.25m/s^2)t

t=92s

displacement=

vf^2=vi^2+2a(dx)

23^2=0^2+2(0.25m/s^2)x =-1058m

Explanation:

you can find time from vf = vi + a(Dt): 0 = 23 m/s + (0.25 m/s/s)t so t = 92 s and you can find the displacement from vf2 = vi2 + 2a(Dx) and find the answer in one step: 232 = 02 + 2(0.25 m/s/s)x so x = -1058 m

6 0

3 years ago