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almond37 [142]
4 years ago
9

Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true abo

ut them as they move apart? (There could be more than one correct choice.)a. Their electrical potential energy keeps decreasing.b. Their acceleration keeps decreasing.c. Their kinetic energy keeps increasing.d. Their kinetic energy keeps decreasing.e. Their electric potential energy keeps increasing.
Physics
1 answer:
Colt1911 [192]4 years ago
7 0

Answer:

Explanation:

correct options

a ) Their electrical potential energy keeps decreasing

Actually as they move apart , their electrical potential energy decreases due to increase of distance between them and kinetic energy increases

so a ) option is correct

b ) Their acceleration keeps decreasing

As they move apart , their mutual force of repulsion decreases due to increase of distance between them so the acceleration decreases .

c ) c. Their kinetic energy keeps increasing

Their kinetic energy increases because their electrical potential energy decreases . Conservation of energy law will apply .

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find the x-component of this vector 18.4,0.250. Remember, angles are measured from the +x axis. Find the x-component and y-compo
Ksivusya [100]

Answer:

x = 0.237

y = 0.0789

Explanation:

Vector with direction 18.4° and magnitude 0.250 has x and y components of:

x = 0.250 cos 18.4°

x = 0.237

y = 0.250 sin 18.4°

y = 0.0789

7 0
3 years ago
Read 2 more answers
The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr
MA_775_DIABLO [31]

Answer:

q=49Q/64

and

x =16R/15

Explanation:

See  attached figure.

E_{Q}= E due to sphere

E_{q}= E due to particule

E_{total}=E_{Q}-E_{q}=0  (1)

according to the law of gauss and superposition Law:

E_{Q}=E_{1}+E_{2}=E_{2} ; electric field due to the small sphere with r1=R/4

E_{Q}=kq_{2}/(r_{1}^{2})=

q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}

then: E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})  (2)

on the other hand, for the particule:

E_{q}=kq/(r_{p}^{2})

r_{p}=2R-R/4=7R/4   ⇒    E_{q}=16kq/(49R^{2})   (3)

We replace (2) y (3) in (1):

E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})

q=49Q/64

--------------------

if R<x<2R   AND E_{total}=E_{Q}-E_{q}=0

E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})

remember that  q=49Q/64

then:

Q(2R-x^{2})=49/64*x^{2}

solving:

x_{1} =16R/15

x_{2} =16R

but: R<x<2R  

so : x =16R/15

7 0
3 years ago
water in a cup and a kettle can have the same temperature even though the quantities are different . give reasons​
jekas [21]

Answer:

The reason is because both are exposed to a virtually infinite heat sink, due to the virtually infinite mass  and of the surrounding environment, compared to the sizes of either the cup or the kettle such that the equilibrium temperature, T_{(equilibrium)} reached is the same for both the cup and the kettle as given by the relation;

\infty M_{(environ)} \times  c_{(environ)} \times (T_2 - T_1) = m_{1} \times  c_{(water)} \times (T_3 - T_2) + m_{2} \times  c_{(water)} \times (T_4 - T_2)

Due to the large heat sink, T₂ - T₁ ≈ 0 such that the temperature of the kettle and that of the cup will both cool to the temperature of the environment

Explanation:

4 0
3 years ago
On an unknown planet in the outer-reaches of the solar system, a pendulum with a 12 g bob and a string length of 4 m oscillates
11Alexandr11 [23.1K]

The acceleration due to gravity (g) on this planet is 39.44 m/s²

<h3>What is solar system?</h3>

Solar system consists of all the planets and the most importantly the center of the solar system is Sun.

Given is an unknown planet in the outer-reaches of the solar system, a pendulum with a 12 g bob and a string length of 4 m oscillates with a period of 2 seconds.

The time period of the pendulum is

T = 2π √l/g

Squaring both sides, we get

l/g = T² / 4π²

g = 4π²l/ T²

Substitute Time period T = 2s and  length l = 4m,  we get

g = 4π²x 4/ 2²

g =39.44 m/s²

Thus, the acceleration due to gravity on this planet is 39.44 m/s²

Learn more about solar system.

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3 0
2 years ago
2. A portion of the body receives 0.15 mGy from radiation with a quality factor Q = 6 and 0.22 mGy from radiation with Q = 10. (
DiKsa [7]

Answer with Explanation:

We are given that

D_1=0.15mGy

D_2=0.22mGy

Q_1=6

Q_2=10

a.We have to find the total dose

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Using the formula then, we get

D=0.15+0.22

D=0.37mGy

b.We have to find the total dose equivalent

Total dose equivalent=H=D_1Q_1+D_2Q_2

Using the formula

H=0.15(6)+0.22(10)

H=3.1mSv

7 0
3 years ago
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