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3 years ago

Match the correct term with the definition

Physics

1 answer:

klio [65]3 years ago

7 0

-Positive particle in the nucleus is a (proton)

-Question two..(isotope)

-Number of protons that an atom has..(atomic number)

-protons + neutrons in the nucleus..(atomic mass)

-neutrally charged element in the nucleus..(neutron)

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Which of the following is most like an indicator of a chemical change

qaws [65]

Answer:

methyl orange, methyl red,phenoptalin, merhy red

Explanation:

all this following are indicators use to check the end point of a reaction

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4 years ago

Using this formula a = F/m What acceleration results from exerting a 125N force on a 0.65kg

Softa [21]

Answer:

Acceleration = 192.3 m/s² (Approx.)

Explanation:

Given:

Force = 125 N

Mass of ball = 0.65 kg

Find:

Acceleration

Computation:

We know that;

Acceleration = Force / Mas

So,

Acceleration = 125 / 0.65

Acceleration = 192.3 m/s² (Approx.)

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3 years ago

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lidiya [134]

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4 years ago

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avanturin [10]

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4 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$