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Vlada [557]
3 years ago
11

I need Help ASAP, and actually help, not just for the points.

Physics
2 answers:
UNO [17]3 years ago
7 0

Answer:

750 force

Explanation:

I have never worked with force but I can guess using the formula. 1500, which is the mass, multiplied by the acceleration, 0.5, would equal 750 force, if being applied by the equation listed, Force= mass×acceleration

Zigmanuir [339]3 years ago
7 0
Force is 7500 1500 multiplied by 5 is 7500
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photoshop1234 [79]

Answer:

8-0!/

Explanation:

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When does a nebula become a star?
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When the contracting gas and dust from nebula become so dense and hot that nuclear fusion starts
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3 years ago
when an object moves down and does not stop which force is acting more strongly on the object, friction or gravity? explain
lina2011 [118]

Gravity acts more strongly on the object.

<u>Explanation:</u>

When an object is dropped from a height, it reaches the ground despite friction acting on it because the force of gravity acting on it is stronger than the air resistance and friction. Air resistance and friction acts upward and prevents the ball from falling. However, it is negligible. The gravity acting on the object is so strong that it pulls the object towards earth with a constant acceleration called as acceleration due to gravity which has a constant value of 9.8m/s².

7 0
3 years ago
The moon Umbriel orbits Uranus (mass = 8.68 x 10^25 kg) at a distance of 2.66 x 10^8 m. What is Umbriel's orbital period (in hou
Whitepunk [10]

Answer:

T = 99.51 hour

Explanation:

Mass of Uranus, M=8.68\times 10^{25}\ kg

The moon Umbriel orbits Uranus at a distance of 2.66\times 10^8\ m

We need to find Umbriel's orbital period. Let it is T. Using Kepler's third law of motion to find it.

T^2\propto r^3\\\\T^2=\dfrac{4\pi^2r^3}{GM}\\\\T^2=\dfrac{4\pi^2\times (2.66\times 10^8)^3}{6.67\times 10^{-11}\times 8.68\times 10^{25}}\\\\T=358244.51\ s

As 1 hour = 3600 s

358244.51 s = 99.51 hour

Hence, Umbriel's orbital period is 99.51 hour.

7 0
2 years ago
The following table lists the work functions of a few commonmetals, measured in electron volts.
steposvetlana [31]

Answer:

Lithium

Explanation:

The equation for the photoelectric effect is

\frac{hc}{\lambda}= \phi + K_{max}

where

\frac{hc}{\lambda} is the energy of the incident photon, with

h being the Planck constant

c is the speed of light

\lambda is the wavelength of the photon

\phi is the work function of the metal (the minimum energy needed to extract the photoelectron from the metal)

K_{max} is the maximum kinetic energy of the emitted photoelectrons

In this problem, we have

\lambda= 190 nm = 1.9\cdot 10^{-7}m is the wavelength of the incident photon

K_{max}=4.0 eV is the maximum kinetic energy of the electrons

First of all we can find the energy of the incident photon

E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.90\cdot 10^{-7} m}=1.05\cdot 10^{-18} J

Converting into electronvolts,

E=\frac{1.05\cdot 10^{-18} J}{1.6\cdot 10^{-19} J/eV}=6.6 eV

So now we can re-arrange the equation of the photoelectric effect to find the work function of the metal

\phi = E-K_{max}=6.6 eV - 4.0 eV=2.6 eV

So the metal is most likely Lithium, which has a work function of 2.5 eV.

3 0
3 years ago
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