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3 years ago

A current of 0.5A flows in a circuit. Determine the quantity of charge that crosses a point in 4 minutes

Physics

1 answer:

zhannawk [14.2K]3 years ago

5 0

0.5 A means 0.5 Coulomb of charge every second.

4 minutes = 240 seconds

(0.5 C/sec) x (240 sec)= 120 Coulombs

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A force of 35 N acts on a ball for 0.2 s. If the ball is initially at rest:

olya-2409 [2.1K]

To solve this problem we will apply the concepts related to momentum and momentum on a body. Both are equivalent values but can be found through different expressions. The impulse is the product of the Force for time while the momentum is the product between the mass and the velocity. The result of these operations yields equivalent units.

PART A ) The Impulse can be calculcated as follows

$L= F\Delta t$

Where,

F = Force

$\Delta t =$ Change in time

Replacing,

$L = (35N)(0.2s)$

$L= 7N\cdot s$

PART B) At the same time the momentum follows the conservation of momentum where:

Initial momentum= Final momentum

And the change in momentum is equal to the Impulse, then

$\Delta p = L$

And

$\Delta p = p_f - p_i$

There is not initial momentum then

$\Delta p = p_f$

$L = p_f$

$p_f = 7N\cdot s = 7kg\cdot m/s$

8 0

3 years ago

In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an

natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

$A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}$

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

$=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa$

σ=5517.25 Pa

Strain in x direction

ε=σ/E

$=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}$

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0

3 years ago

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3

Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

$\tau=\sigma\cos\alpha\cos\beta$

Put the value into the formula

$\tau=12\cos60\times\cos35$

$\tau=4.91\ MPa$

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

$\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}$

Put the value into the formula

$\sigma=\dfrac{6.2}{\cos60\cos35}$

$\sigma=15.13\ MPa$

Hence, This is the required solution.

3 0

3 years ago

An airplane has wings, each with area A, designed so that air flows over the top of the wing at 265 m/s and underneath the wing

exis [7]

Answer

given,

Pressure on the top wing = 265 m/s

speed of underneath wings = 234 m/s

mass of the airplane = 7.2 × 10³ kg

density of air = 1.29 kg/m³

using Bernoulli's equation

$P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2$

$\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P =\dfrac{1}{2}\times 1.29\times (265^2-234^2)$

$\Delta P =9977.5 Pa$

Applying newtons second law

2 Δ P x A - mg = 0

$A =\dfrac{mg}{2\Delta P}$

$A =\dfrac{7.2\times 10^3 \times 9.8}{2\times 9977.5}$

A = 3.53 m²

7 0

3 years ago

True or False? PLEASE HELP ME

viva [34]

Answer:

a. True - Joules is the unit measure for energy.

b. False - Potential energy is associated with position

c. False - Kinetic energy is associated with movement.

d. False - It's climbing, which means it also has kinetic energy.

6 0

3 years ago