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sergeinik [125]
3 years ago
6

Why are most objects not magnetic?

Physics
2 answers:
Vedmedyk [2.9K]3 years ago
6 0
Because there’s like no metal stuff idk
maksim [4K]3 years ago
6 0

Answer:if the magnetic fields of all those atoms are randomly oriented then they would cancel each other out and the material would have no magnetic field

Explanation:

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A ball is released from a hot air balloon moving downward with a velocity of -10.0 meters/second and a height of 1,000 meters. H
Nikolay [14]
Here, ball is released... and it is in free fall means with zero initial velocity.

We know, s = ut + 1/2 at²
Here, s = 1000 m
u = 0
a = 10 m/s2

Substitute their values, 
1000 = 0 + 1/2 * 10 * t²
2000 = 10 * t²
t² = 2000 /10
t = √200
t = 14.14 s

In short, Your Answer would be 14.14 seconds

Hope this helps!
7 0
3 years ago
Read 2 more answers
Nathalie leaves a history classroom and walks 3 meters North to drinking fountain. Then she turns and walks 10 meters south to a
pishuonlain [190]

Answer:

13 meters

Explanation:

Step one:

given

We are told that Nathalie leaves a history classroom and walks 3 meters North

Then travels another 10 meters south to an art classroom.

Required

The total distance.

Step two:

The total distance can be computed by summing up the 3 meter North distance traveled and the 10 meter south distance traveled

Total distance= 3+10= 13meters

7 0
3 years ago
The gravity of Earth is attracting a person towards the center with 500N of gravitational force. The person is exerting a reacti
mestny [16]

Answer:

500n

Explanation:

7 0
3 years ago
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Tides are a result of
Virty [35]
The answer should be D but if its not then im sorrry, idk
6 0
3 years ago
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The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it
Lilit [14]

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, m=0.02 kg

Kinetic energy imparted, K=1200 J

Length of rifle barrel, d=1 m

(a)

Let the speed of bullet when it leaves the barrel is v.

Kinetic energy, K=\frac{1}{2} mv^{2}

v=\sqrt{\frac{2K}{m} }

=\sqrt{\frac{2\times1200}{0.02} }

=346.4m/s

Initial speed of bullet, u=0

The average speed in the barrel, v_a_v_g=\frac{u+v}{2}

=\frac{0+346.4}{2} \\=173.2 m/s

Time taken by bullet to cross the barrel, t=\frac{d}{v}

=\frac{1}{173.2}\\ =0.00577 second

Power, P_a_v_g=\frac{W}{t}

=\frac{1200}{0.00577} \\=207.97kW

(b)

In projectile motion,

Maximum height, H_m=\frac{v^2\sin^2\theta}{2g} \\

Range, R=\frac{v^2\sin2\theta}{g}

given that, H_m=R

then, \frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter

5 0
3 years ago
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