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KatRina [158]
2 years ago
6

I WILL GIVE BRAINLIEST

Physics
2 answers:
frez [133]2 years ago
3 0
D. Magnet attracting metal.
This is because magnetic forces are non contact forces; they pull or push on objects without touching them.
harina [27]2 years ago
3 0

Answer:

d, magnetic attracting metal, because the force is still going on without the magnet touching metal, and you cant see the force

Explanation:

Hope this helps <33

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An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see
mash [69]

Answer:

(7.90 × 10⁻¹⁵) J

Explanation:

The electric force exerted on the elecrron by rhe electric field is given by

F = qE

where |q| = charge on the particle = (1.602 × 10⁻¹⁹) C

E = magnitude of the electric field = (2.9 × 10⁶) V/m or N/C

F = 1.602 × 10⁻¹⁹ × 2.9 × 10⁶ = (4.646 × 10⁻¹³) N

From Newton's first law of motion relation, we can obtain the acceleration this force confers on the electron

F = ma

m = mass of the electron = (9.11 × 10⁻³¹) kg

a = acceleration of the electron caused by the electric force = ?

(4.646 × 10⁻¹³) = (9.11 × 10⁻³¹) × a

a = (4.646 × 10⁻¹³)/(9.11 × 10⁻³¹)

a = (5.10 × 10¹⁷) m/s²

Now, using the equations of motion, we can obtain the velocity with which the electron reaches the positive plate

u = initial velocity of the electron = 0 m/s (since the electron was initially at rest)

v = final velocity of the electron = ?

a = acceleration of the electron = (5.10 × 10¹⁷) m/s²

y = distance covered by the electron = 1.7 cm = 0.017 m

v² = u² + 2ay

v² = 0² + 2(5.10 × 10¹⁷)(0.017)

v² = (1.734 × 10¹⁶)

v = 131,677,182.5 m/s = (1.32 × 10⁸) m/s

Kinetic energy with which the electron hits the positive plate = (1/2)(m)(v²) = (1/2)(9.11 × 10⁻³¹)(1.32 × 10⁸)² = (7.90 × 10⁻¹⁵) J

Hope this Helps!!!

3 0
3 years ago
Which of the following is the only requirement for an object to be in projectile motion? A. The horizontal and vertical motions
NARA [144]
The answer is d, gravity is the only force acting on the object
8 0
3 years ago
A 2.0-kg ball rolls to the right at 3.0 m/s. A 4.0-kg ball rolls to the left at 2.0 m/s . What is the momentum of the system aft
charle [14.2K]

Answer:

Final momentum after a head on collision is -2kgm/s

Explanation:

         One ball moves to the right and the other moves opposite  and momentum is a vector quantity so that considering the direction

Initial momenta are        P₁=2x3=6kgm/s        P₂=4x(-2)=-8kgm/s      

Final momentum is the vector sum of P(final)= 6-8= -2 kgm/s

4 0
3 years ago
Starting from your campsite you walk 3.0 km east, 6.0 km north, 1.0 km east, and then 4.0 km west. How far are you from your cam
Hatshy [7]
Think of it like a graph. You start at the origin which is (0,0).  go three to the east which now you are (3,0). Then, six to the north. Now, you are at (3,6).  1 to the east, ((4,6).  Then you go 4 to the west which is back tracking. So, you end at (0,6) which is saying you are now 6 km north from your campsite. 

Hope this helps!
6 0
3 years ago
When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

\alpha = Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

3 0
3 years ago
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