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3 years ago

“What are three ways that a cat pushing on a cat-flap door can change the amount of torque applied to the door?”

1 answer:

7 0

It depends on where the cat is applying the force and how much the force is .. after all where and at what is distance from the axis of rotation of the door

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A proton with a speed of 3.2 x 106 m/s is shot into a region between two plates that are separated by a distance of 0.23 m. As t

harina [27]

Answer:

<h2>The magnitude of the magnetic is 0.145 T</h2>

Explanation:

Given :

Speed of proton $v = 3.2 \times 10^{6}\frac{m}{s}$

Mass of proton $m = 1.67 \times 10^{-27}$ Kg

The force on the proton in magnetic field is given by,

$F = q (v \times B)$

$F = qvB \sin \theta$

But $\sin 90 = 1$ (∵ Force is perpendicular to the velocity so $\theta = 90$ )

$F = qvB$

When particle enter in magnetic field at the angle of 90° so particle moves in circle

So force is given by,

$F = \frac{mv^{2} }{r}$

Where $r =$ radius but in our case 0.23 m, $q = 1.6 \times 10^{-19}$ C

By comparing above two equation,

$B = \frac{mv}{qr}$

$B = \frac{1.67 \times 10^{-27} \times 3.2 \times 10^{6} }{1.6 \times 10^{-19} \times 0.23 }$

$B = 0.145$ T

5 0

2 years ago

What type of material is wrapped around an object could cause the object to lose the most heat

Dmitrij [34]

<span>A sheet of copper could cause the object to lose the most amount of heat. Copper is an essential element and a good conductor of heat. Heat can transfer from one end of a piece of copper to the other end.</span>

7 0

2 years ago

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with

Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = $\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2$

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 31 J

Final KE = $\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J$

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0

3 years ago

One of the world's largest Ferris wheels, the Cosmo Clock 21 with a radius of 50.0 m is located in Yokohama City, Japan. Each of

STatiana [176]

Answer:

a = 0.55 m / s²

Explanation:

The centripetal acceleration is given by the relation

a = v² / r

angular and linear velocities are related

v = w r

we substitute

a = w² r

In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system

w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s

let's calculate

a = 0.105² 50.0

a = 0.55 m / s²

4 0

3 years ago

A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of

Nesterboy [21]

Answer:

The common velocity v after collision is 2.8m/s²

Explanation:

look at the attachment above ☝️

3 0

2 years ago