1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Andrews [41]
3 years ago
10

A push on a textbook with 10 N moves it to a distance of 2 cm. How much force is need to move it to 4 cm?

Physics
1 answer:
uysha [10]3 years ago
8 0

Answer:

20N

Explanation:

Ratio of N to cm-

10:2

so to make 2=4 times 2 so The ratio is now-

20:4

so to move 4 cm you need to push 20N.

You might be interested in
A cat climbs 10 m directly up a tree.
Harman [31]
 <span>there is no horizontal displacement if he went straight up 

straight up means vertical, so his vertical displacment is 20 m</span>
6 0
3 years ago
A student claims that the Chernobyl accident proves that nuclear power is on safe for use in the United States which statement i
Setler79 [48]
Hi! Check out my valid counter argument below!

"The accident only released harmless gamma rays."

Hope I helped!
5 0
3 years ago
A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a
Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0
3 years ago
A horizontal compass is placed 21 cm due south from a straight vertical wire carrying a 36 a current downward. in what direction
Anit [1.1K]

 <span>
The needle of a compass will always lies along the magnetic field lines of the earth. 
A magnetic declination at a point on the earth’s surface equal to zero implies that 
the horizontal component of the earth’s magnetic field line at that specific point lies along 
the line of the north-south magnetic poles. </span>

The presence of a current-carrying wire creates an additional <span>
magnetic field that combines with the earth’s magnetic field. Since magnetic 
<span>fields are vector quantities, therefore the magnetic field of the earth and the magnetic field of the vertical wire must be combined vectorially. </span></span>

<span>
Where:</span>

B1 = magnetic field of the earth along the x-axis = 0.45 × 10 ⁻ ⁴ T

B2 = magnetic field due to the straight vertical wire along the y-axis

We can calculate for B2 using Amperes Law:

B2 = μ₀ i / [ 2 π R ]

B2 = [ 4π × 10 ⁻ ⁷ T • m / A ] ( 36 A ) / [ 2 π (0.21 m ) ] <span>
B2 = 5.97 × 10 ⁻ ⁵ T = 0.60 × 10 ⁻ ⁴ T </span>

The angle can be calculated using tan function:<span>
tan θ = y / x = B₂ / B₁ = 0.60 × 10 ⁻ ⁴ T / 0.45 × 10 ⁻ ⁴ T <span>
tan θ = 1.326</span></span>

θ = 53°

<span>
<span>The compass needle points along the direction of 53° west of north.</span></span>

8 0
3 years ago
A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
irinina [24]

288.51 N is  the magnitude of the force that the beam exerts on the hi.nge.

Given

Mass 0f beam = 40 Kg

The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N

Angle between the beam and cable is = 90°

Angle between beam and the horizontal component = 31°

As the system of the beam, hi_nge and cable are in equilibrium.

The magnitude of the force that the beam exerts on the hi_nge can be calculated by -

F =The  horizontal component of force + the vertical component of force  

F = 86.62 N + 40 × 9.8 × sin 31°

F =86.62 N + 201.89 N

F = 288.51 N

Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51  N.

Learn more about components of forces here brainly.com/question/26446720

#SPJ1

7 0
2 years ago
Other questions:
  • A particle of charge +13 µC and mass 3.57 10-5 kg is released from rest in a region where there is a constant electric field of
    14·1 answer
  • Why can’t contour lines cross?
    8·1 answer
  • ________ Layer of the atmosphere where jets fly through and contains ozone layer
    10·1 answer
  • Which of the following is a chemical reaction? A. A spoon bending B. Wood burning C. Plates being stacked D. A book being opened
    15·2 answers
  • The reactivity of an atom arises primarily from ___.
    9·1 answer
  • A 2.81 μF capacitor is charged to 1220 V and a 6.61 μF capacitor is charged to 560 V. These capacitors are then disconnected fro
    12·1 answer
  • Must all species reproduce in order for the species to survive?
    9·2 answers
  • A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.
    5·1 answer
  • Who could explain the phenomenom of interference of light?​
    5·1 answer
  • Meteorologists are interested in the relationship between minimum pressure and maximum wind speed
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!