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Sonbull [250]
3 years ago
13

When you push a 1.85-kg book resting on a

Physics
1 answer:
Margaret [11]3 years ago
6 0

Answer:

Given:

mb=2.05 kgmb=2.05 kg Mass of the book

fs=2.45 Nfs=2.45 N Static friction

fk=1.50 Nfk=1.50 N Kinetic friction

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Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p
Marta_Voda [28]

Answer:

Acceleration, a=9.91\times 10^{15}\ m/s^2

Explanation:

It is given that,

Separation between the protons, r=3.73\ nm=3.73\times 10^{-9}\ m

Charge on protons, q=1.6\times 10^{-19}\ C

Mass of protons, m=1.67\times 10^{-27}\ kg

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

ma=\dfrac{kq^2}{r^2}

a=\dfrac{kq^2}{mr^2}      

a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}      

a=9.91\times 10^{15}\ m/s^2

So, the acceleration of two isolated protons is 9.91\times 10^{15}\ m/s^2. Hence, this is the required solution.

3 0
3 years ago
If the initial velocity of an object was -2 meters per second
Shalnov [3]
A :-) for this question , we should apply
a = v - u by t
Given - u = -2 m/s
v = -10 m/s
t = 16 sec
Solution -
a = v - u by t
a = -10 - -2 by 16
a = -12 by 16
( cut 12 and 16 because 2 x 6 = 12 and
2 x 8 = 16 )
( cut 6 and 8 because 2 x 3 = 6 and
2 x 4 = 8 )
a = 3 by 4
a = 0.75 m/s^2

.:. The acceleration is 0.75 m/s^2.
3 0
3 years ago
Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter yo
Tems11 [23]

Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

Therefore, we can calculate the power using above formula

P= 1/5.70 + 1/0.02

Power=50.17dioptre

Therefore, the power the eye's is using to see the object from distance is 5.70D

3 0
4 years ago
On Ramesh’s13th birthday, his father invited all his friends and their relatives. It was a big party with lots of food and DJs.
BaLLatris [955]
I go inside and take a nap normally that will help maybe play music
3 0
3 years ago
A car is traveling down the highway cruise control at 50m/s (this means its is traveling with a constant velocity NOT accelerati
lawyer [7]

Given parameters:

Speed of car = 50m/s

Time of travel = 5minutes (300s)

Unknown;

Distance of travel = ?

To solve this problem, we need to understand how speed relates with distance and time.

 Speed is a physical quantity. It is the distance traveled divided by time taken.

                Speed  = \frac{distance}{time}

   So,

                Distance  =  speed x time

Input the parameters and solve for the distance;

               Distance  = 50m/s x 300s

               Distance  = 15000m

The distance covered during this time interval and speed is 15000m

3 0
3 years ago
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