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erastova [34]
3 years ago
6

Here is a picture showing how two magnets will stick to one another when opposite poles align. When these magnets are pulled sli

ghtly apart, does the attractive force between them still exist?
A)Yes, but the new attractive force is due to gravity.
B)No, since they are not touching, no force is at work.
C)Yes, even though they are not touching, they are pulling on one another.
D)No, when the magnets are apart, they exert a repulsive force on one another.
Physics
2 answers:
7nadin3 [17]3 years ago
7 0

Answer:

C

Explanation:

A magnetic field exerts its force beyond just direct touch.

ser-zykov [4K]3 years ago
7 0
The correct answer is C!
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A man pushes a heavy box. if the man exerts a force of 200 N on the box to keep it moving at a constant velocity what is the fri
Ad libitum [116K]

Answer:

Kinectic friction is the answer to the lower part of the question

5 0
3 years ago
An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens
Pavlova-9 [17]

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

\frac{1}{v} -\frac{1}{u} =\frac{1}{f}

Putting the values

\frac{1}{v} +\frac{1}{15} =\frac{1}{14}

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

\frac{210}{15} \times \frac{D}{f_e}

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

7 0
3 years ago
A wheel, starting from rest, rotates with a constant angular acceleration of 1.80 rad/s^2. During a certain 7.00 s interval, it
lora16 [44]

Answer:

a) 1.3 rad/s

b) 0.722 s

Explanation:

Given

Initial velocity, ω = 0 rad/s

Angular acceleration of the wheel, α = 1.8 rad/s²

using equations of angular motion, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

where

θ2 - θ1 = 53.2 rad

t2 - t1 = 7s

substituting these in the equation, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²

53.2 = 7.ω(0) + 1/2 * 1.8 * 49

53.2 = 7.ω(0) + 44.1

7.ω(0) = 53.2 - 44.1

ω(0) = 9.1 / 7

ω(0) = 1.3 rad/s

Using another of the equations of angular motion, we have

ω(0) = ω(i) + α*t1

1.3 = 0 + 1.8 * t1

1.3 = 1.8 * t1

t1 = 1.3/1.8

t1 = 0.722 s

5 0
3 years ago
An airplane is cruising at a speed of 250 m/s. If the airplanes engines provide a forward force of 19,540 N, calculate the force
ExtremeBDS [4]
250/300 dived by 500
3 0
3 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 44 ft/s2. What is the dist
elena55 [62]

Answer:

Distance, d = 61.13 ft

Explanation:

It is given that,

Initial speed of the car, u = 50 mi/h = 73.34 ft/s

Finally, it stops i.e. v = 0

Deceleration of the car, a=-44\ ft/s^2

We need to find the distance covered before the car comes to a stop. Let the distance is s. It can be calculated using third law of motion as :

v^2-u^2=2as

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{0-(73.34\ ft/s)^2}{2\times -44\ ft/s^2}

s = 61.13 ft

So, the distance covered by the car before it comes to rest is 61.13 ft. Hence, this is the required solution.

4 0
4 years ago
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