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MArishka [77]
3 years ago
10

At the maximum height of a projectile’s trajectory—

Physics
2 answers:
Marina CMI [18]3 years ago
7 0

Well, a, c, and d are all true, but the simple reason is that the velocity is zero at the top of the flight. The best, simplest answer is 'a'.

uranmaximum [27]3 years ago
5 0

D. Cross product of velocity & amp; acceleration is 0

You might be interested in
Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
3 years ago
The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2
Darya [45]

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

i = magnitude of current

magnitude of current is given as

i = \frac{Ane^{2}\tau E}{m}

i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}

i  = 1.87 A

4 0
3 years ago
An electron experiences a magnetic force with a magnitude of 4.90×10−15 n when moving at an angle of 60.0 ∘ with respect to a ma
Leya [2.2K]
Missing questions: "find the speed of the electron".

Solution:
the magnetic force experienced by a charged particle in a magnetic field is given by
F=qvB \sin \theta
where
q is the particle charge
v its velocity
B the magnitude of the magnetic field
\theta the angle between the directions of v and B.

Re-arranging the formula, we find:
v= \frac{F}{qB \sin \theta}
and by substituting the data of the problem (the charge of the electron is q=1.6 \cdot 10^{-19} C), we find the velocity of the electron:
v= \frac{F}{qB \sin \theta}= \frac{4.90 \cdot 10^{-15}N}{(1.6 \cdot 10^{-19}C)(3.70 \cdot 10^{-3} T)(\sin 60^{\circ})}=9.56 \cdot 10^6 m/s
4 0
3 years ago
Given one mole of diamond vs one mole of graphite,
grandymaker [24]

Answer:

The pressure is P= -  6.39*10^8Pa

The temperature is T =1218.63 K

Explanation:

Generally Gibbs free energy is mathematically represented as

                   G = E + PV -TS

   Where  E is the enthalpy

               PV is the pressure volume energy (i.e PV energy)

                S  is the entropy

                T is the temperature

For stability to occur the Gibbs free energy must be equal to zero

Considering Diamond

  So at temperature of  T = 300 K

         E + PV - TS = 0

making P the subject

          P = \frac{TS-E}{V}

Now substituting 300 K for T , 2900 J  for E ,

                              3.42cm^3 = \frac{3.42}{1*10^6} = 3.42*10^{-6}m^3 for V and 2.38 J/K for S

     P = \frac{(300 * 2.38)- 2900}{3.42*10^{-6}}

         P= -  6.39*10^8Pa

The negative sign signifies the direction of the pressure

Given that  P = 1*0^5Pa

making T the subject

            T = \frac{PV+E}{S}

Substituting into the equation

            T = \frac{1*10^5 * 3.42 *10^{-6}+2900}{2.38}

                T =1218.63 K

             

         

7 0
3 years ago
Read 2 more answers
A 1.0-m long wire is carrying a certain amount of current. The wire is placed perpendicular to a magnetic field of strength 0.20
Marysya12 [62]

For a current-carrying wire running perpendicular to a magnetic field, the magnetic force acting on the wire is given by:

F = ILB

F = magnetic force, I = current, L = wire length, B = magnetic field strength

Given values:

F = 0.60N, L = 1.0m, B = 0.20T

Plug in and solve for I:

0.60 = I(1.0)(0.20)

I = 3.0A

5 0
3 years ago
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