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Paha777 [63]
3 years ago
5

Alfred pushes on a heavy box, but cannot move it. The box has a lot of _____. inertia motion friction gravity

Physics
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

Inertia

Explanation:

Inertia is best defined as the ability of an object to resist a change in position or movement. That is why when an object has a higher mass, the higher the inertia. Imagine an oncoming truck that is fully loaded versus you. The tendency for the truck to change its movement would be difficult because of its its mass. It has a lot of inertia.

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8. Una fuerza de 100 N actúa sobre un cuerpo que se desplaza a lo largo de
vitfil [10]

Answer:

50m

Explanation:

8 0
4 years ago
Which statement describes the use of control rods?
Anit [1.1K]
Control rods<span> are </span>rods<span>, plates, or tubes containing a neutron absorbing material (material with high absorbtion cross-section for thermal neutron) such as boron, hafnium, cadmium, etc., used to </span>control<span> the power of a nuclear reactor.
so the statement that describe a control rod is:
</span><span>Control rods are lowered into the reactor to slow down the reaction by absorbing neutrons.</span>
6 0
4 years ago
Read 2 more answers
Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o
IrinaVladis [17]

Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

F_{att}=G\frac{m_{1}m_{2}}{r^{2}}

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N

Force of attraction between 435 kg mass and 38 kg mass is

F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N

Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

8 0
4 years ago
A high-speed railway car goes around a flat, horizontal circle of radius 480 m at a constant speed. The magnitudes of the horizo
kvv77 [185]

Answer:

Net force ( F ) = 208 N.

Speed of car ( v ) = 68.58 m/s.

Explanation:-

The radius ( r ) = 480 m.

Mass of passenger ( m ) = 65 kg.

Horizontal force ( f_{x} ) = 208 N.

Vertical force ( f_{n} ) = 637 N.

Net vertical component of force ( f_{x} ) = m*g - 637   = 65 * 9.8 - 637 =  0 N.

a ) magnitude of net force .

F = \sqrt{f^{2} _{x} + f^{2} _{y}  }   = \sqrt{208^{2} + 0^{2} }   = 208 N.

b) The speed of car.

f_{n}  = \frac{m*v^{2} }{r}    where v is the velocity.

637 = \frac{65 *v^{2} }{480}

v² = 4704

v = 68.58 m/s.

6 0
3 years ago
Yung's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a sc
Lady bird [3.3K]

Answer:

1.082 mm

Explanation:

From the question, we can see that we were given The following

Wavelength of the atoms, λ = 502 nm = 502*10^-9 m

Radius of the screen away from the double slit, r = 1.1 m

We know that Y(20) = 10.2 mm = 10.2*10^-3 m

d = (20 * R * λ) / Y(20)

d = (20 * 1.1 * 502*10^-9)/10.2*10^-3

d = 1.1*10^-5 / 10.2*10^-3

d = 1.082 mm

Therefore, we can say that the distance of separation between the two slits is 1.082 mm

4 0
4 years ago
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