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4 years ago

When an objects mass increases its gravitational pull ____ a.decreases b.increases c. Stays the same

Physics

1 answer:

Lemur [1.5K]4 years ago

8 0

I believe that an object's gravitational pull would increase the heavier it gets.

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How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will

nordsb [41]

Answer:

m = 1 kg

Explanation:

Given that,

The force constant of the spring, k = 39.5 N/m

The frequency of oscillation, f = 1 Hz

The frequency of oscillation is given by the formula as formula as follows :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f^2=\dfrac{k}{4m\pi^2}\\\\m=\dfrac{k}{4\pi^2 f^2}\\\\m=\dfrac{39.5}{4\pi^2 \times (1)^2}\\\\m=1\ kg$

So, the mass that is attached to the spring is 1 kg.

6 0

3 years ago

Calculate the force exerted on the wall assuming that force is horizontal and using the data in the schematic representation of

Jlenok [28]

Answer:

1.93 x 10∧3 N

Explanation:

The picture attached shows the calculation

8 0

3 years ago

Are graded receptor potentials always depolarizing? Do graded receptor potentials always make it easier to induce action potenti

Oksana_A [137]

Answer: Yes,graded receptor potential always depolarize.

Yes,graded receptor potentials must occur to depolarize the neutrons to threshold before action potentials can occur.

Explanation:

7 0

4 years ago

A laser pulse of duration 25 ms has a total energy of 1.4 J. The wavelength of this radiation is

SpyIntel [72]

Answer:

n = 4 x 10¹⁸ photons

Explanation:

First, we will calculate the energy of one photon in the radiation:

$E = \frac{hc}{\lambda}\\\\$

where,

E = Energy of one photon = ?

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of radiation = 567 nm = 5.67 x 10⁻⁷ m

Therefore,

$E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{5.67\ x\ 10^{-7}\ m}$

E = 3.505 x 10⁻¹⁹ J

Now, the number of photons to make up the total energy can be calculated as follows:

$Total\ Energy = nE\\1.4\ J = n(3.505\ x\ 10^{-19}\ J)\\n = \frac{1.4\ J}{3.505\ x\ 10^{-19}\ J}\\$

<u>n = 4 x 10¹⁸ photons</u>

8 0

3 years ago

Solve 3* +5-220t = 0

slega [8]

Answer:

t = 27.5

Explanation:

$3 + 5 -220t = 0$

Well to solve for t we need to combine like terms and seperate t.

So 3+5= 8

8 - 220t = 0

We do +220 to both sides

8 = 220t

And now we divide 220 by 8 which is 27.5

Hence, t = 27.5

4 0

3 years ago