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LUCKY_DIMON [66]
2 years ago
7

What are the similarities between the compass needle (magnetism) and a charge to probe the electric field (electricity)?

Physics
1 answer:
puteri [66]2 years ago
6 0

Similarities between the compass needle (magnetism) and a charge to probe the electric field (electricity) is that Electric fields are produced by two kinds of charges, positive and negative similarly Magnetic fields are associated with two magnetic poles

Magnetic Field is the region around a magnetic material or a moving electric charge within which the force of magnetism acts. A pictorial representation of the magnetic field which describes how a magnetic force is distributed within and around a magnetic material.

Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

Electric fields are produced by two kinds of charges, positive and negative. Magnetic fields are associated with two magnetic poles, north and south, although they are also produced by charges (but moving charges).

To learn more about electric fields here

brainly.com/question/15800304

#SPJ4

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When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees
nexus9112 [7]

(a) The man comes to a halt in 1.99 ms = 1.99 × 10⁻³ s, so his average acceleration slowing him from 6.33 m/s to a rest is

<em>a</em> = (6.33 m/s - 0) / (1.99 × 10⁻³ s) ≈ 3180 m/s²

so that the average net force on the man during the landing is

<em>F</em> = (70.8 kg) <em>a</em> ≈ 225,000 N

i.e. with magnitude 225,000 N.

(b) With knees bent, the man has an average acceleration of

<em>a</em> = (6.33 m/s - 0) / (0.147 s) ≈ 43.1 m/s²

and hence an average net force of

<em>F</em> = (70.8 kg) <em>a</em> ≈ 3050 N

(c) The net force on the man is

∑ <em>F</em> = <em>n</em> - <em>w</em> = <em>m a</em>

where

<em>n</em> = magnitude of the normal force, i.e. the force of the ground pushing up on the man

<em>w</em> = the man's weight, <em>m g</em> ≈ 694 N

<em>m</em> = the man's mass, 70.8 kg

<em>g</em> = mag. of the acceleration due to gravity, 9.80 m/s²

<em>a</em> = the man's acceleration

Using the acceleration in part (b), we have

<em>n</em> = <em>m g</em> + <em>m a</em> = <em>m</em> (<em>g</em> + <em>a</em>)

<em>n</em> = (70.8 kg) (9.80 m/s² + 43.1 m/s²) ≈ 3740 N

4 0
3 years ago
Which is not a characteristic of a heat pump?
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Describe an object that emits radiation in Astronomy
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Answer:

the sun

Explanation:

the sun emits all types of electromagnetic radiation but a vast majority of it comes in the for of in the for of visible light ultraviolet rays and infrarays

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If two vectors are 7 cm and 5 cm .then what's the addition of these two?
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Answer:

7 + 5 = 12

7 + 5 = 12    

Explanation:

Since it asks for 2 vectors, you have to add 2 times.

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3 years ago
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la
Nookie1986 [14]

Answer:

h=53.09m         (2)

v_{min}>5.05m/s

v_{max}

Explanation:

<u>a)Kinematics equation for the first ball:</u>

v(t)=v_{o}-g*t

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

y_{o}=h       initial position is the building height

v_{o}=8.9m/s      

The ball reaches the ground, y=0, at t=t1:

0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}

h=1/2*g*t_{1}^{2}-v_{o}t_{1}           (1)

Kinematics equation for the second ball:

v(t)=v_{o}-g*t

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

y_{o}=h       initial position is the building height

v_{o}=0       the ball is dropped

The ball reaches the ground, y=0, at t=t2:

0=h-1/2*g*t_{2}^{2}

h=1/2*g*t_{2}^{2}         (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03              (3)

We solve the equations (1) (2) (3):

1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}

g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)

g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)

-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)

2.06*gt_{1}-2v_{o}t_{1}=g*1.06

t_{1}=g*1.06/(2.06*g-2v_{o})

vo=8.9m/s

t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s

t2=t1-1.03              (3)

t2=3.29sg

h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m         (2)

b)t_{1}=g*1.06/(2.06*g-2v_{o})

t1 must :   t1>1.03  and t1>0

limit case: t1>1.03:

1.03>9.81*1.06/(2.06*g-2v_{o})

1.03*(2.06*9.81-2v_{o})

20.8-2.06v_{o}

(20.8-10.4)/2.06

v_{min}>5.05m/s

limit case: t1>0:

g*1.06/(2.06*g-2v_{o})>0

2.06*g-2v_{o}>0

v_{o}

v_{max}

8 0
4 years ago
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