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3 years ago

OF BRICK (a) The pushing force does not make the brick move. Explain why.

Physics

1 answer:

Sever21 [200]3 years ago

6 0

Every force has an equal but opposite reaction, the forces are balanced out on the brick.

Weight (mg) acts downwards from the brick but the normal contact force from the table is acting upwards therefore resultant force is 0N equilibrium

You know the direction of the force is based upon the resultant force, so if you were to push the brick horizontally across the table it would travel horizontally

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Why is the process of photosynthesis important to food webs?

Slav-nsk [51]

Answer:

to know what that specif animal or thing does in the food chain

Explanation:

5 0

3 years ago

In the 2016 Olympics in Rio, after the 50 m freestyle competition, a problem with the pool was found. In lane 1 there was a gent

Dmitry_Shevchenko [17]

Answer:

Explanation:

A )

speed of swimming in still water is given by the expression

distance / time

= 50 / 25

= 2 m /s

In lane 1 , 1.2 cm/s current is flowing in the direction that the swimmers are going so swimmers will cover distance at the rate of 2 + 1.2 = 3.2 m /s.

time to cover distance of 50 m in lane 1

= distance / speed

= 50 / 3.2 = 15.625 s

In lane 8 , 1.2 cm/s current is flowing against the direction that the swimmers are going so swimmers will cover distance at the rate of 2 - 1.2 = .8 m /s.

time to cover distance of 50 m in lane 1

= distance / speed

= 50 / .8 = 62.5 s

8 0

3 years ago

The acceleration due to gravity on the Moon's surface is

Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

$g'=\frac{g}{6}$

We know that weight of an object on Earth is,

$w = m\times g$

$m = \frac{w}{g}$

Similarly, weight on Moon will be

$w' = m\times g'$

$w' = \frac{w}{g}\times\frac{g}{6}$

$w' = \frac{300}{6}$

$w' = 50$

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

7 0

3 years ago

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3 0

2 years ago

Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25

mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp = 1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0

3 years ago