answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nikolay [14]
1 year ago
12

The total negative charge on the electrons in 1kg of helium (atomic number 2, molar mass 4) is____________.

Physics
2 answers:
tekilochka [14]1 year ago
4 0

Answer:

Explanation:

n = \frac{m}{M}

n = \frac{1000}{4}

         = 250 moles.

    N  = n×6.02×10^{23}

        = 1.505×10^{26}

Total charge = (1.505×10^{26}) × (1.6×10^{-19})

                     = 2.4×10^{7} C.

vazorg [7]1 year ago
3 0

Answer:

Number of electrons=2

charge of an electron, e=1.6*10^-19

Q=ne     Q=2*1.6*10^-

-19

m=1kg=1000g

n=m/M=1000/4=250

N=nL

N=2*1.6*10^-19*250*6.02*10^23

N=4.8*10^7 C

Explanation:

You might be interested in
You hang different masses M from the lower end of a vertical spring and measure the period T for each value of M. You use Excel
Svetradugi [14.3K]

Answer:

a)693.821N/m

b)17.5g

Explanation:

We the Period T we can find the constant k,

That is

T = 2 \pi \sqrt{\frac{m}{k}}

squaring on both sides,

T^2=\frac{4\pi^2}{k}M +\frac{4\pi^2}{k}m_{spring}

where,

M=hanging mass, m = spring mass,

k =spring constant

T =time period

a) So for the equation we can compare, that is,

y=T^2=0.0569x+0.0010

the hanging mass M is x here, so comparing the equation we know that

\frac{4\pi^2}{k}=0.0569\\k= \frac{4\pi^2}{0.0569}\\k=693.821N/m

b) In order to find the mass of the spring we make similar process, so comparing,

\frac{4\pi^2}{k}m =0.001\\m=\frac{0.004k}{4\pi^2} =\frac{0.001*693.821}{4\pi^2}\\m=0.0175kg\\m=17.5g

3 0
2 years ago
In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it start
Hitman42 [59]

Answer:

v_f = 16.6 m/s

Explanation:

As we know by force equation that force along the inclined planed due to gravity is given as

F_g = mg sin\theta

so the acceleration due to gravity along the plane is given as

a = \frac{F_g}{m}

now we have

a = g sin\theta

a = (9.81 sin4.0)

a = 0.68 m/s^2

now we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 9.2^2 = 2(0.68)(140)

v_f = 16.6 m/s

4 0
2 years ago
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
1 year ago
If the loss of 3500 kcal is equal to a loss of 1.0 lb, how many days will it take charles to lose 5.0 lb
rusak2 [61]
We are missing an important piece of information needed to answer this question: the number of kcal Charles losses per day. However, we can come up with a general equation in which kcal/day is the only independent variable.

We know that it takes 3500 kcal to lose one pound. To lose 5 pounds, Charles needs to lose 5 x 3500 kcal = 17,500 kcal. 

To find how many days it takes Charles to lose 17,500 kcal (5 pounds), we must divide that amount by the number of kcal Charles loses per day. 
Here is the equation to calculate that number

Number of days= 17500 / (kcal per day)

If given calories, remember that 1000 calories = 1 kcal, and .001 kcal = 1 cal

4 0
1 year ago
Assume that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic inte
faltersainse [42]

Answer:

Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all - option A

Explanation:

Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all; the reason being that: no magnetic field is being produced by a charge that is static. Only a moving charge can produce a magnetic effect. And the magnet can not have any torque due to its own magnetic lines of force.

5 0
1 year ago
Other questions:
  • Four students measured the acceleration of gravity. The accepted value for their location is 9.78 m/s2. Which student's measurem
    9·2 answers
  • A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth
    14·2 answers
  • Which of the following statements is false?
    6·2 answers
  • When the voltage and current have _____ polarities in a pure capacitive circuit, the capacitor is discharging and the energy is
    9·1 answer
  • A wave with an amplitude of 9.3 mm is traveling along a string whose linear mass density is 230 g/m and whose tension is 65 N. I
    7·1 answer
  • A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
    9·1 answer
  • An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined
    8·1 answer
  • An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu
    10·1 answer
  • The dwarf planet praamzius is estimated to have a diameter of about 300km and orbits the sun at a distance of 6.4E12m . What is
    8·1 answer
  • Question 5 At 12:00 pm, a spaceship is at position ⎡⎣324⎤⎦ km ⎣ ⎢ ⎡ ​ 3 2 4 ​ ⎦ ⎥ ⎤ ​ km away from the origin with respect to so
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!