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babymother [125]
3 years ago
10

How does gravitational force and electric force compare?

Physics
1 answer:
makvit [3.9K]3 years ago
3 0
Both equations show an inverse square relationship between force<span> and separation distance. And both equations show that the </span>force<span> is proportional to the product of the quantity that causes the </span>force<span> 
</span><span>Felect = k • </span>Q1<span> • </span>Q<span>2 </span><span>/ d2</span>

<span>Felect = (</span>9.0 x 109 N•m2/C2<span>) • (1.00 C) • (1.00 C) / (1.00 m)2</span>

<span>Felect = 9.0 x 109 N</span>

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In this experiment, you will use a track, a toy car, and some washers to explore Newton’s first two laws of motion. You will mak
polet [3.4K]

How can we experimentally verify newton's laws?

7 0
3 years ago
Read 2 more answers
2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
  • M is magnification.

{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
4 0
2 years ago
A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction
zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, F_c=\frac{mv^2}{R}

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

f=\mu N

As there is no vertical motion, therefore, N=mg. So,

f=\mu mg

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816

Therefore, option (c) is correct.

7 0
3 years ago
Strontium−90 is one of the products of the fission of uranium−235. This strontium isotope is radioactive, with a half-life of 28
Ludmilka [50]

Answer : The time passed in years is 20.7 years.

Explanation :

Half-life = 28.1 years

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{28.1\text{ years}}

k=2.47\times 10^{-2}\text{ years}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 2.47\times 10^{-2}\text{ years}^{-1}

t = time passed by the sample  = ?

a = initial amount of the reactant  = 1.00 g

a - x = amount left after decay process = 0.600 g

Now put all the given values in above equation, we get

t=\frac{2.303}{2.47\times 10^{-2}}\log\frac{1.00}{0.600}

t=20.7\text{ years}

Therefore, the time passed in years is 20.7 years.

8 0
3 years ago
How much momentum does a 77 kg football player have if he is running with a speed of 4 m/s
worty [1.4K]

Answer:

308 N-s

Explanation:

Momentum is given by

P= mv

P= 77 x 4

P= 308

4 0
3 years ago
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