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BARSIC [14]
3 years ago
13

Help plisssssss:((((

Physics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

i dont know

Explanation:

heheheheheehehsry

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What is the net force on this object?​
Sergio039 [100]
200N

Explanation:
600N-400N = 200N
6 0
3 years ago
How do terrestrial and giant planets differ? List as many ways as you can think of.
Leni [432]

Answer: The differences between terrestrial planets and the giant planets are s follows-

  1. The inner planets namely Mercury, Venus, Earth and Mars are the terrestrial planets, whereas the outer planets namely Jupiter, Saturn, Uranus and Neptune are known as the outer planets.
  2. Inner planets are composed mainly of silicate materials as well as some metals, whereas the giant planets are comprised of water (in different states) and gases such as Hydrogen and Helium.
  3. The density of the inner planets are more in comparison to the outer planets, where earth has the highest of about 5.5 gm/cm³ and Saturn has the lowest of about 0.7 gm/cm³.
  4. Due to the location of the inner planets near to the sun, they have high boiling point, whereas outer planets are much far from the sun so they have a low boiling point.
5 0
3 years ago
Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30
beks73 [17]

Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

J=-D\frac{\Delta C}{\Delta x}

\Delta C is the rate in concentration

\Delta xis the rate in thickness

D is the diffusion coefficient, where,

D= D_0 exp(\frac{Q_d}{RT})

Replacing D in the first law,

J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}

clearing T,

T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}

Replacing our values

T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}

T=-\frac{80000}{-138.09}

T=575.16K

4 0
3 years ago
A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)
wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

7 0
3 years ago
A ramp is 1.0 m high and 3.0 m long. What is the IMA of the ramp?
oksano4ka [1.4K]
To calculate the ideal mechanical advantage for an inclined plane, divide th length of the incline by the height of the incline. 
Therefore; IMA = L/h
L= 3.0 m, while h =1.0 m
 IMA = 3/1
         = 3
Therefore the IMA of the ramp is 3 
This means the ramp increases the force that is being exerted by 3 times.
4 0
3 years ago
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