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4 years ago

5 The frequencies of AM radio waves,

Physics

1 answer:

aleksklad [387]4 years ago

5 0

Message me back and I can help you.

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You are loading a refrigerator weighing 2267 N onto a truck, using a wheeled cart. The refrigerator is raised 1.09 m to the truc

meriva

Answer:

a).

Wmin= 2471.03 J

W=1603.01 J

Explanation:

Weight, w= 2267 N

$w= m*g\\W=m*g*h \\W=w*h$

Minimum work 'h' is the distance the refrigerator is raised h=1.09m

$W_{min}=2267 N* 1.09m\\W_{min}=2471.03 J$

The motion is no frictional force so, the magnitude of the force with a angle of 45.0° is find using:

$W=m*g*h'\\h'= sin(45)\\W=2267N*sin(45)\\W=1603.01 J$

6 0

3 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

3 years ago

If you designed a rollercoaster, how might you design it? Would you have friction?

kvasek [131]

Yes, because you would need friction to slow down the rollercoaster to a stop.

4 0

4 years ago

Please help!!!!

Dafna11 [192]

The intensity of the electric field is 30,000 N/C

Explanation:

The strength of the electric field produced by a single-point charge is given by the equation

$E=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

$q=3\cdot 10^{-9}C$ is the magnitude of the charge

r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity

Substituting, we find:

$E=(8.99\cdot 10^9)\frac{3\cdot 10^{-9}}{(0.03)^2}=30,000 N/C$

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

3 years ago

What is true of both gravity and magnetism?

stira [4]

Explanation:

I want to say option B - Both forces can act without objects touching.

5 0

3 years ago