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3 years ago

15

Providing that it does not collide with another object, what is the most likely reason that an object moving through space would

change velocity?

1 answer:

Korolek [52]3 years ago

8 0

The object might change velocity because it might of combusted or somehow changed the weight

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A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in

Troyanec [42]

Answer:

The value is $\epsilon = 3.84 *10^{-5} \ V$

Explanation:

From the question we are told that

The diameter of the ring is $d = 18 \ mm = 0.018 \ m$

The length of the solenoid is $l = 25 \ cm = 0.25 \ m$

The diameter of the solenoid is $D = 5.0 \ cm = 0.05 \ m$

The number of turns is N = 1500

The change in current in the solenoid is $\Delta I = 20 \ A$

The time taken is $\Delta t = 1 \ s$

Generally the radius of the ring is

$r = \frac{d}{2}$

=> $r = \frac{0.018 }{2}$

=> $r = 0.009 \ m$

Generally the area of the ring is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * 0.009^2$

=> $A = 2.545 *10^{-4}\ m^2$

Generally the induced emf is mathematically represented as

$\epsilon = A * \frac{dB}{dt}$

Here

$\frac{dB }{dt} = \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi *10^{-7} \ N/A^2$

So

$\frac{dB }{dt} = 4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}$

=> $\frac{dB }{dt} = 0.150816\ T/s$

So

$\epsilon = 0.150816 * 2.545 *10^{-4}$

=> $\epsilon = 3.84 *10^{-5} \ V$

3 0

3 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Scrat [10]

Answer:

5.24 m/s

Explanation:

So for the rod to be able to rise upward to the straight up position, the kinetic energy caused by linear speed v0 must be just enough to convert into the potential energy.

Since the rod is uniform in mass, we can treat the body as 1 point, at its center of mass, or geometric center, aka 0.35 / 2 = 0.175 m from the pivot.

For the rod to swing from bottom to top, the center must have moved a distance of h = 0.175 * 2 = 0.35 m, vertically speaking.

Since we neglect friction and air resistance, according to the law of energy conservation then:

$E_k = E_p$

$mv^2/2 = mgh$

Where v is the speed at the center of mass, g = 9.81 m/s2 is the gravitational acceleration, and m is the mass. We can divide both sides by m

$v^2 = 2gh = 2*9.81*0.35 = 6.867$

$v = \sqrt{6.867} = 2.62 m/s$

As this is only the speed at the center of mass, the speed at the bottom end would be different, to calculate this, we need to find the common angular speed:

$\omega = v / r = 2.62 / 0.175 = 14.97 rad/s$

Where r is the rotation radius, or the distance from pivot point to the center of mass

$v_0 = \omega R = 14.97*0.35 = 5.24 m/s$

Where R is the distance from the pivot to the bottom end of the rod

5 0

3 years ago

What is the mass of a bicycle that has 90 kg.m/s of momentum at a speed of 6 m/s?

OLEGan [10]

Answer:

m = 15 kg

Explanation:

p = m × v

90 = m × 6

6m = 90

m = 90/6

m = 15 kg

6 0

2 years ago

The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =

luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = $\frac{\textup{Distance}}{\textup{Velocity}}$

or

Time = $\frac{\textup{0.016}}{\textup{23}}$

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = $\frac{q\times8.20\times10^4}{1\times10^{-11}}$

Now from the Newton's equation of motion

$d=ut+\frac{1}{2}at^2$

where,

d is the distance

u is the initial speed

a is the acceleration

t is the time

or

$0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2$

or

q = 9.98 × 10⁻⁹ C

4 0

3 years ago

5. Calculate the acceleration of a 2 kg block across a table if you push with a force of 20 N and the frictional force is 4 N.

frozen [14]

Answer:

Explanation:hey do you go to GOC too?

8 0

2 years ago