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8090 [49]
3 years ago
11

6. Jim wishes to push a 100. N wood crate across a wood floor.What is the minimum horizontal force that would be required to sta

rt the crate moving?
Physics
2 answers:
sergejj [24]3 years ago
7 0
I want to say the answer is 42N
Usimov [2.4K]3 years ago
5 0

Answer:

50 N

Explanation:

The minimum horizontal force required to start the crate moving must be equal to the maximum force of static friction that acts between the floor and the crate.

The maximum force of static friction is given by:

F_f = \mu N

where

\mu is the coefficient of static friction

N is the normal reaction exerted by the floor on the crate.

For wood-wood, as in this case, the maximum coefficient of static friction is about \mu=0.5.

For a horizontal surface, the normal reaction is equal to the weight of the object, so in this case N=100 N. Therefore, the minimum horizontal force required will be:

F=F_f = \mu N=(0.5)(100 N)=50 N

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You see lightning and 30 seconds later you hear thunder. how far away is the thunderstorm? take the speed of sound to be 339 m/s
Jobisdone [24]
Let the observer be 'd' distance away from the thunderstorm and let light take 't' time to reach the observer
Since the speed of sound and light remains constant in a particular medium, we can use
      Speed = Distance/Time

For light,
   3 x 10^8 = d/t
                t = d/(3 x 10^8)   -1 

For sound,
           339 = d/(t + 30)       -2

Putting value from 1 in 2.
               d = 10^4 m(approx)
3 0
3 years ago
I will give u BRAINILIEST and 15 points
o-na [289]
The answer is D interferometry
3 0
3 years ago
Read 2 more answers
after the car leaves the platform , gravity causes it to accelerate downward at a rate of 9.8 m/s2. what is the gravitational fo
alexandr1967 [171]

The gravitational force on the car is the force popularly known
as the car's "weight".  Its magnitude is

            (9.8 m/s²) times (the car's mass, in kilograms) .

The unit of this quantity is [newton] .

5 0
3 years ago
Questions -
irga5000 [103]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of  X =  1000 \  N and  Y  =  1400 \ N  

Explanation:

Considering the diagram,

   Taking moment about the pivot , we have

           X *  1 =  400 *  2.5

=>         X =  1000 \  N

Generally given that the bar is at equilibrium then the upward forces is equal to the down ward force

So

        Y  =  X  +  400

=>     Y  =  1000  +  400

=>     Y  =  1400 \ N    

   

3 0
2 years ago
An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the
Montano1993 [528]

Answer:

The speed of the object just before it hits Earth is  \sqrt{\dfrac{GM}{R}}

(A) is correct option.

Explanation:

Given that,

M = mass of earth

R = radius of earth

The potential energy at height above the surface of the earth

P.E=-\dfrac{GmM}{R+h}

The kinetic energy at height above the surface of the earth

K.E = 0

The total energy at height above the surface of the earth

E = K.E+P.E

E = -\dfrac{GmM}{R+h}....(I)

The total energy at the surface of the earth

E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}....(II)

We need to calculate the speed of the object  just before it hits Earth

From equation (I) and (II)

-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}

Here, h = R

-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}

v= \sqrt{\dfrac{GM}{R}}

Hence, The speed of the object just before it hits Earth is  \sqrt{\dfrac{GM}{R}}.

4 0
3 years ago
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