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Irina-Kira [14]
3 years ago
11

What is a property of “normal force”? a. It always points perpendicular to the contact surface. b. It always points parallel to

the contact surface. c. It always points up. d. It always completely counters gravity.
Physics
1 answer:
OleMash [197]3 years ago
6 0

Answer:

a. It always points perpendicular to the contact surface.

Explanation:

"Normal" means perpendicular.  Normal forces are always perpendicular to the contact surface.

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Can you guys please help me on this one?​
LenKa [72]

If the object is not at rest how?

4 0
2 years ago
Read 2 more answers
A cement block accidentally falls from rest from the ledge of a 52.9-m-high building. When the block is 14.3 m above the ground,
Anna007 [38]

Answer:

The man has at most 0.418 secs to get out of the way

Explanation:

To determine how much time at most the man has to get out of the way, we will calculate the time it will take the block to reach height 1.94m from height 14.3m.

To do this, we will first determine the time it will take the block to reach height 1.94 m from height 52.9 m and find the time it takes the block to reach height 14.3m above the ground from the same height (52.9 m), the difference is the time the man has to get out of the way.

Now, the time it will take the block to reach height 1.94 m from height 52.9 m

This means the time it will take the block to travel a height distance of 52.9m - 1.94m = 50.96m

From one of the equations of motions for free falling bodies

h = ut + 1/2(gt²)

Where h is the height

u is the initial velocity

t is the time

and g is the acceleration due to gravity (Take g = 9.8 m/s²)

From the question, the block falls from rest

∴ u = 0 m/s

h = 50.96 m

Putting these into the equation

50.96 = 0(t) + 1/2(9.8)(t²)

50.96 = 4.9t²

t² = 50.96/4.9

t² = 10.4

t = √10.4

t = 3.225 secs

This is the time it will take to reach height 1.94m (that is to reach the man)

For the time it takes the block to reach height 14.3m above the ground from height 52.9 m

That is, the time it takes the block to travel a height distance of 52.9m - 14.3m = 38.6 m

Here,

h = 38.6 m

and u = 0 m/s

Putting these into the same equation

h = ut + 1/2(gt²)

38.6 = 0(t) + 1/2(9.8)(t²)

38.6 = 4.9t²

t² = 38.6/4.9

t² = 7.878

t = √7.878

t = 2.807 secs

This is the time it takes the block to reach height 14.3 m

Now, the difference in time is 3.225secs - 2.807 secs = 0.418 secs

Hence, the man has at most 0.418 secs to get out of the way.

7 0
3 years ago
A dolphin can swim at a constant speed of 12.5 m/s. How
myrzilka [38]

Answer:

\boxed {\tt 3.6 \ seconds}

Explanation:

Time can be found by dividing the distance by the speed.

t=\frac{d}{s}

The distance is 45 meters and the speed is 12.5 meters per second.

d= 45 \ m \\s= 12.5 \ m/s

t=\frac{45 \ m}{12.5 \ m/s}

Divide. Note that the meters, or "m" will cancel each other out.

t=\frac{45 }{12.5 \ s}

t=3.6 \ s

It will take the dolphin 3.6 seconds to swim a distance of 45 meters are 12.5 meters per second.

6 0
3 years ago
Read 2 more answers
What is one property of iodine
mamaluj [8]

Answer: I2

Explanation:

6 0
3 years ago
(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit
Ainat [17]

Answer:

a

The number of fringe is  z  = 3 fringes

b

The  ratio is I = 0.2545I_o

Explanation:

a

 From the question we are told that

        The wavelength is  \lambda = 600 nm

        The distance between the slit is  d = 0.117mm = 0.117 *10^{-3} m

        The width of the slit is  a = 35.7 \mu m = 35.7 *10^{-6}m

let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

             z = \frac{d}{a}

Substituting values

             z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}  

             z  = 3 fringes

b

   From the question  we are told that the order  of the bright fringe is  n = 3

   Generally the intensity of  a pattern  is mathematically represented as

                 I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]

Where I_o is the intensity  of the  central fringe

 And  Generally  sin \theta = \frac{n \lambda }{d}

               I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]

               I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]

               I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]

                I = I_o (1)(0.2545)

                  I = 0.2545I_o

6 0
3 years ago
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