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sveta [45]
3 years ago
5

Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.

Physics
1 answer:
Rufina [12.5K]3 years ago
4 0

Answer: 1. decreasing the mass of both objects

2. decreasing the mass of one of the objects

3. increasing the distance between the objects

Explanation: Hope that helped! (:

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A beam of photons with wavelength 150 nm and beam of electrons having the same energy as the photons go through the same slit of
satela [25.4K]

Answer:

Explanation:

For moving particle , the wavelength is given by the expression

\lambda=\frac{h}{\sqrt{2mE} }

E is kinetic energy of the particle

Here particle is electron and its kinetic energy

E = 150 nm

= 1244 / 150

= 8.3 eV

8.3 X 1.6 X 10⁻¹⁹ J

=13.28 X 10⁻¹⁹ J

Putting these values in the expression for wavelength as given above we have

\lambda=\frac{6.6\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times13.28\times10^{-19}} }

λ = .425 nm

Wave length of electron wave will be 0.425 nm

Hence its wavelength will be smaller and therefore magnitude of angle made by it will also be smaller. So magnitude of β will be smaller than that of α .

4 0
3 years ago
A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit
djverab [1.8K]

Answer:

The current is  I  = 6.68 \  A

Explanation:

From the question we are told that  

     The radius of the loop is  r =  6 \ cm  = 0.06 \ m

     The  earth's magnetic field is B_e =  0.7G=  0.7  G * \frac{1*10^{-4} T}{1 G}  = 0.7 *10^{-4} T

      The  number of turns is  N  =1

Generally the magnetic field generated by the current in the loop is mathematically represented as

        B  =  \frac{\mu_o  * N  *  I}{2 r }

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

         B  =  B_e

=>     B_e =  \frac{\mu_o  *  N  *  I  }{ 2 * r}

     Where  \mu is the permeability of free space with value  \mu _o  =   4\pi * 10^{-7} N/A^2

       0.7  *10^{-4}=  \frac{ 4\pi * 10^{-7}  * 1 * I}{2 * 0.06}

=>     I  =  \frac{2 *  0.06 *  0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}

       I  = 6.68 \  A

3 0
4 years ago
2 (a) A plane from airport A flies 280 km to the east to airport B. The plane then travelled north to airport C, 190 km away. (i
Sonbull [250]

Answer:

See below

Explanation:

280 km   east then 190 km north

Use Pythagorean theorem to find the resultant displacement

d^2 = 280^2 + 190^2

d = 338.4 km

Angle will be    arctan ( 190/280) = 34.16 °

8 0
2 years ago
A hockey puck attached to a horizontal spring oscillates on a frictionless, horizontal surface. The spring has force constant 4.
bekas [8.4K]

Answer:

m = 0.164 kg

Explanation:

T (period)

k (force/spring constant)

m (mass)

T = 2*Pi*sqrt(m/k)

T/(2*Pi) = sqrt(m)/sqrt(k)

(T/(2*Pi))*sqrt(k) = sqrt(m)

m = ((T/(2*Pi))*sqrt(k))^2

m = 4.5*((1.2/(2*Pi)))^2

m = 0.1641403175

3 0
3 years ago
Solve the below problems being sure to provide the correct significant figures.
morpeh [17]


Sum of the first ten digits is:
7+3+4+5+4+1+7+8+0 = 39

39, when divided by 9 give you the remainder of 3

9 x4, is 36
36 +3 equals 39
(39 = 9 x 4 + 3)

So X= 0
5 0
4 years ago
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