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laiz [17]
3 years ago
9

When a system absorbs heat from its surroundings it is a(n) ____ process

Physics
1 answer:
Nezavi [6.7K]3 years ago
8 0
C. endothermic

An endothermic process takes heat from the surroundings while an exothermic process gives out heat to the surroundings.


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maw [93]

Answer:mountains

Explanation:

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The mass of an atom is
suter [353]

Answer:

Atomic mass is defined as the number of protons and neutrons in an atom, where each proton and neutron has a mass of approximately 1 amu (1.0073 and 1.0087, respectively). The electrons within an atom are so miniscule compared to protons and neutrons that their mass is negligible.

I hope this is the answer you were looking for :D

4 0
3 years ago
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The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5t, collide with molecules of t
andreev551 [17]
The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.

The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.

When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.

a)

Magentic force, F = q*v*B

q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T

Centripetal force, F = m*Ac = m * v^2 / R

where,

Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit

Now equal the two forces: q*v*B = m * v^2 / R => R =  m*v / (q*B)

=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]

=> R = 0.114 m

b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.

R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]

=> R = 10.4 m

 

4 0
4 years ago
A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou
zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius  = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}

E=6.88\times10^{7}\ N/C

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d).  For, r = 7.00 cm

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}

E=5.43\times10^{6}\ N/C

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0
3 years ago
A particle starts from the origin at t = 0 with an initial velocity of 4.8 m/s along the positive x axis.If the acceleration is
tia_tia [17]

Answer:

Part a)

Velocity = 6.9 m/s

Part b)

Position = (3.6 m, 5.175 m)

Explanation:

Initial position of the particle is ORIGIN

also it initial speed is along +X direction given as

v_x = 4.8 m/s

now the acceleration is given as

\vec a = -3.2 \hat i + 4.6 \hat j

when particle reaches to its maximum x coordinate then its velocity in x direction will become zero

so we will have

v_f = v_i + at

0 = 4.8 - 3.2 t

t = 1.5 s

Part a)

the velocity of the particle at this moment in Y direction is given as

v_f_y = v_i + at

v_f_y = 0 + 4.6(1.5)

v_f_y = 6.9 m/s

Part b)

X coordinate of the particle at this time

x = v_x t + \frac{1}{2}a_x t^2

x = 4.8(1.5) - \frac{1}{2}(3.2)(1.5^2)

x = 3.6 m

Y coordinate of the particle at this time

y = v_y t + \frac{1}{2}a_y t^2

y = 0(1.5) + \frac{1}{2}(4.6)(1.5^2)

y = +5.175 m

so position is given as (3.6 m, 5.175 m)

5 0
4 years ago
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