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dsp73
3 years ago
11

on the surface of planet x, a body with a mass of 10 kilograms weighs 40. newtons. the magnitude of the acceleration due to grav

ity on the surface of planet X is

Physics
1 answer:
VARVARA [1.3K]3 years ago
7 0
Based on the Newton's second law of motion, the value of the net force acting on the object is equal to the product of the mass and the acceleration due to gravity. If we let a be the acceleration due to gravity, the equation that would allow us to calculate it's value is,

      W = m x a

where W is weight, m is mass, and a is acceleration. Substituting the known values,

    40 kg m/s² = (10 kg) x a

Calculating for the value of a from the equation will give us an answer equal to 4. 

ANSWER: 4 m/s². 
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horsena [70]

Answer:

Explanation:

velocity=frequency*wavelength

velocity = 2.99*10^8 m/s

frequency = ?

wavelength = 3.012*10^-12

2.99*10^8m/s = (f)(3.012*10^-12)

f=9.58*10^19 Hertz    

5 0
3 years ago
A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

  • If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
  • So, we can write the following equation:

       E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)

  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
  • So, on the outer surface of the shell there must be a charge that be the difference between them:

        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

  • As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
6 0
3 years ago
Help me asap its due today
Margaret [11]

Answer:

3.54* 10^{22} N

Explanation:

Using the formula you gave:

F_g = \frac{6.67*10^{-11}*2.0*10^{30}*5.97^{24}  }{(1.5*10^{11})^2 }

3 0
3 years ago
Please I need help on Physics ASAP!!!
Dafna11 [192]

Answer:

you cant cheat on test

Explanation:

you will get repoted

7 0
3 years ago
Which of the following statements are true? Positively charged objects attract other positively charged objects. Negatively char
butalik [34]
The statement that is true is that positively charged objects attract negatively charged objects. This is due to a law that states 'like forces attract while unlike forces repel. This same concept applies to magnetism. If you put two similar poles together, for example; if you place two south poles together. You feel a separating force between the two poles. But if you place two opposite poles together they attract each other. Hope i helped. <span />
7 0
3 years ago
Read 2 more answers
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