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maksim [4K]
3 years ago
14

The value of the normal force exerted on a 2,000 gram block on a table

Physics
1 answer:
kobusy [5.1K]3 years ago
4 0

Answer:

N = 19.6 N

Explanation:

Given that,

Mass of a block, m = 2000 g

1 kg = 1000 g

It means, 2000 g = 2kg

We need to find the value of normal force on the block on a table. Normal force is balanced by the weight of the block as follows :

N = mg, g is acceleration due to gravity

N = 2 kg × 9.8 m/s²

N = 19.6 N

So, the normal force acting on the block is 19.6 N.

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A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x ax
alexdok [17]

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

so

applying conservation law of momentum

p + .444 i + .748 j  = 0

p = - .444 i -  .748 j  

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

.024 v = .87

v = 36.25 m / s .

6 0
3 years ago
An atom with seven valence electrons would most likely lose an electron to become stable true or fasle
mel-nik [20]

Answer:

false

Explanation:

since it has seven valence electrons it means it's a non metal and non metals always gain electrons.

5 0
2 years ago
Over a 24-hour period, the tide in a harbor can be modeled by one period of a sinusoidal function. the tide measures 5.15 ft at
RSB [31]
<span>f(x) = 5.05*sin(x*pi/12) + 5.15

   First, you need to determine the period of the function. The period will be the time interval between identical points on the sinusoidal function. For this problem, the tide is rising and at 5.15 at midnight for two consecutive days. So the period is 24 hours. Over that 24 hour period, we want the parameter passed to sine to range from 0 to 2*pi. So the scale factor for x will be 2*pi/24 = pi/12 which is approximately 0.261799388. The next thing to note is the magnitude of the wave. That will simply be the difference between the maximum and minimum values. So 10.2 ft - 0.1 ft = 10.1 ft. And since the value of sine ranges from -1 to 1, we need to divide that magnitude by 2, so 10.1 ft / 2 = 5.05 ft.

   So our function at this point looks like f(x) = 5.05*sin(x*pi/12) But the above function ranges in value from -5.05 to 5.05. So we need to add a bias to it in order to make the low value equal to 0.1. So 0.1 = X - 5.05, 0.1 + 5.05 = X, 5.15 = X. So our function now looks like:
  f(x) = 5.05*sin(x*pi/12) + 5.15

   The final thing that might have been needed would have been a phase correction. With this problem, we don't need a phase correction since at X = 0 (midnight), the value of X*pi/12 = 0, and the sine of 0 is 0, so the value of the equation is 5.15 which matches the given value of 5.15. But if the problem had been slightly different and the height of the tide at midnight has been something like 7 feet, then we would have had to calculate a phase shift value for the function and add that constant to the parameter being passed into sine, making the function look like:
 f(x) = 5.05*sin(x*pi/12 + C) + 5.15
  where
 C = Phase correction offset.

   But we don't need it for this problem, so the answer is:
 f(x) = 5.05*sin(x*pi/12) + 5.15

   Note: The above solution assumes that angles are being measured in radians. If you're using degrees, then instead of multiplying x by 2*pi/24 = pi/12, you need to multiply by 360/24 = 15 instead, giving f(x) = 5.05*sin(x*15) + 5.15</span>
7 0
3 years ago
Hi gir_ls join nkd-mbja-nuj​
GREYUIT [131]

Answer:

never lol

studying is your work

but why all are doing I don't know=_=

3 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
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