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3 years ago

12

The direction of the acceleration of an object on a(n) _ path is toward the _ of the

circle.

1 answer:

aleksandrvk [35]3 years ago

8 0

Answer:

circular...center

Explanation:

Physics

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An atom that has a negative four charge will have which of the following? Question 1 options: 4 more protons than electrons An e

Lilit [14]

Electrons: negative charge
Protons: positive charge
Neutrons: negative charge

The atom would have to have more electrons than protons

Hope this helps :)

3 0

2 years ago

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

Semenov [28]

Answer:

$f_n=3.75N$

Explanation:

From the question we are told that:

Frictional force $F=0.150N$

Coefficient of kinetic friction $\mu=0.04$

Generally the equation for Normal for is mathematically given by

$f_n=\frac{F}{\mu}$

Therefore

$f_n=\frac{0.150}{0.04}$

$f_n=3.75N$

5 0

2 years ago

A physical change is a change to the physical properties of a substance. This type of change does

LiRa [457]

Answer:

False

Explanation:

5 0

2 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

8 0

2 years ago

The magnitude of electrical force between a pair of charged particles is ____ as much when the particles are moved half as far a

Gnesinka [82]

The magnitude of the electrical force between a pair of charged particles is 4 Times as much when the particles are moved half as far apart.

This can be easily understood by Columb's law,

$F_{new} = \frac{kQ_{1}Q_{2}}{r^{2}}$

which state's that the amount of electrical force experienced by two charged particles is inversely proportional to the square of the distance between them.

∴ $\frac{F_{new} }{F_{old} } = \frac{Distance_{new}^{2} }{Distance_{old}^{2} }$

Now, we know the new distance is half the original distance,

$F_{new} = \frac{kQ_{1}Q_{2}}{\frac{r}{2}^{2} } \\F_{new} = 4\frac{kQ_{1}Q_{2}}{r^{2}}$

$F_{new} = 4F_{old}$

The electrical force of attraction or electrostatic force of attraction between two charged particles refers to the amount of attractive or repulsive force that exists between the two charges. This can be calculated by Columb's Law.

A charged particle in physics is a particle that has an electric charge. It might be an ion, such as a molecule or atom having an excess or shortage of electrons in comparison to protons. The same charge is thought to be shared by an electron, a proton, or another primary particle.

Learn more about electrical force here

brainly.com/question/2526815

#SPJ4

8 0

1 year ago