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3 years ago

As an ice cube melts, its molecules

Physics

1 answer:

fomenos3 years ago

3 0

Answer:

A. absorb heat energy and move farther apart

Explanation:

The answer is A since when a ice cube takes in heat, it changes form and melts down to a liquid : water.

The molecules in water are farther apart since they are moving around alot but can be contained in a container.

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A 790kg car moving at 7 m/s takes a turn around a circle with a radius of 20 m. Determined the net force (in Newton’s) acting up

prisoha [69]

1935.5 N is the "net force" acting on a car.

<u>Explanation</u>:

Given that,

Mass of the car is 790 kg.

Velocity of the car is 7 m/s. (v)

It turned around with 20 m. (r)

We know that, Net force = m × a

$\text { Here, acceleration of the car is radial acceleration } a_{\mathrm{rad}}=\frac{v^{2}}{r}$

$\mathrm{a}_{\mathrm{rad}}=\frac{7^{2}}{20}$

$\mathrm{a}_{\mathrm{rad}}=\frac{49}{20}$

$a_{\text {rad }}=2.45 \mathrm{m} / \mathrm{s}^{2}$

Now, Net force = m × a

Net force = 790 × 2.45

Net force = 1935.5 N

4 0

3 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

3 years ago

Which of these surface features are formed because of the heating of groundwater?

Tju [1.3M]

The correct option is C.
Hot springs and geysers are formed as a result of been heated up by heat from the interior of the earth. This type of heat is called geothermal heat. A hot spring refers to a spring that is produced as a result of geothermally heated water. Hot springs usually have very high temperature. Geysers is a form of hot spring and it refers to a pool of water that has seeped through a opening in the earth surface.

8 0

3 years ago

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 1.00 mm away. part a

mihalych1998 [28]

<span>We can use Coulomb's law to find the force F acting on the proton that is released. F = k x Q1 x Q2 / r^2 k = 9 x 10^9 Q1 is the charge on one proton which is 1.6 x 10^{-19} C Q2 is the same charge on the other proton r is the distance between the protons F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2 F = 2.304 x 10^{-22} N We can use the force to find the acceleration. F = ma a = F / m a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg) a = 1.38 x 10^5 m/s^2 The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>

8 0

3 years ago

Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest an

SpyIntel [72]

Answer:

a) 2.43 m/s

b) 4.83 m/s

c) 0.023 m/s²

Explanation:

a) Both cars cover a distance of 510 m in 210 s. Since car A has no acceleration

Speed = Distance / Time

$\text{Speed}=\frac{510}{210}=2.43\ m/s$

Velocity of car A is 2.43 m/s

t = Time taken = 210 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 510 m

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 510=0\times 210+\frac{1}{2}\times a\times 210^2\\\Rightarrow a=\frac{510\times 2}{210^2}\\\Rightarrow a=0.023\ m/s^2$

Acceleration of car B is 0.023 m/s²

$v=u+at\\\Rightarrow v=0+0.023\times 210\\\Rightarrow v=4.83\ m/s$

Final velocity of car B is 4.83 m/s

6 0

3 years ago