A 0.2 oz. bullet leaves the muzzle of a rifle with a speed of 1420 ft/s. If the length of the barrel is 24 inches, what is the magnitude of the force acting on the bullet while it travels down the barrel

Answer:

thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.

Answer:

nuclear fusion

Answer:

nuclear fusion

Explanation:

The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom.

Answer:

P V = N R T       ideal gas equation

V1 = k * T1        if P is constant and also N and R will be constant

V2 = k * T2      where k is some constant

Or     V2 = (T2 / T1) * V1      also known as "Charles Law"  for expansion at    

 constant pressure

V2 = (373 / 273) * 1000 cm^3 = 1366 cm^3     where T is absolute temperature

Answer:Answer:

Initial value problem is:

u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0

u'(0) = 0.03m/s

Explanation:

The directions of Fd(t*) and U'(t*) are not specified in the question, so we'll take Fd(t*) to be negative and U'(t*) to be positive. This is due to the fact that the damping factor acts in the direction opposite the direction of the motion of the mass.

M = 5kg; L= 10cm or 0.1m;

F(t) = 10 sin(t/2) N ; Fd(t*) = - 2N

U'(t*) = 4cm/s or 0.04m/s

u(0) = 0

u'(0) = 3cm/s or 0.03m/s

Now, we know that W = KL.

Where K is the spring constant.

And L is the length of extension.

So, k = W/L

W= mg = 5 x 9.81 = 49.05N

So,k = 49.05/0.1 = 490.5kg/s^(2)

Now from spring damping, we know that; Fd(t*) = - γu'(t*)

Where,γ = damping coefficient

So, γ = - Fd(t*)/u'(t*)

So, γ = 2/0.04 = 50 Ns/m

Therefore, the initial value problem which describes the motion of the mass is;

5u'' + 50u' + 490u = (10 sin(t/2) N

Divide each term by 5 to give;

u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0

u'(0) = 0.03m/s

Explanation:

Answer:

hi your pinterest I'd or Twitter I'd pls.

Answer:

your answer is -788231

Explanation:

1+1677=1678

1678-789909=-788231

Answer:

(a) 14.14 s

(b) 4.24 s

Explanation:

maximum acceleration, a =  1 m/s^2

maximum speed, v = 5 m/s

initial speed, u = 0 m/s

(a) distance, s = 100 m

Let the time is t.

Use second equation of motion

[tex]s = u t 0.5 at^2\\\\100 = 0 + 0.5 \times 1 \times t^2\\\\t = 14.14 s[/tex]

(b) distance, s = 9 m

Let the time is t'.

Use second equation of motion

[tex]s = u t + 0.5 at^2\\\\9= 0 + 0.5 \times 1 \times t'^2\\\\t' = 4.24 s[/tex]

Answer:

The first person will gain more velocity as a result of recoil.

Explanation:

Let us recall that from Newton's third law of motion, action and reaction are equation and opposite. A consequence of this law is the proposition that ''momentum can neither be created nor destroyed.''

Hence, when two people who have the same mass, throw two different objects at the same velocity but the first object is heavier than the second, the first object possesses greater momentum than the second object hence the first person will gain more velocity as a result of recoil.

Answer:

V = 29.49 m/s

Explanation:

Given that,

The mass of a car,[tex]m_c=2100\ kg[/tex]

The mass of a motorcycle, [tex]m_m=290\ kg[/tex]

The initial velocity of the car,[tex]v_c=30i-10j[/tex]

[tex]|v_c|=\sqrt{30^2+(-10)^2} =31.62\ m/s[/tex]

The initial velocity of the motorcycle,[tex]v_m=10i+10j[/tex]

[tex]|v_m|=\sqrt{10^2+10^2} =14.14\ m/s[/tex]

As they stick together. Let V is the speed. So, using the conservation of momentum,

[tex]m_cv_c+m_mv_m=(m_c+m_m)V\\\\V=\dfrac{m_cv_c+m_mv_m}{(m_c+m_m)}\\\\V=\dfrac{2100\times 31.62+290\times 14.14}{(2100+290)}\\\\V=29.49\ m/s[/tex]

So, the velocity of the stuck together car and the motorcycle after the collision is 29.49 m/s.

[tex]A. \: \: \: \: F = kqq/r^ 2 \\B. \: \: \: \: F = kqq/r \\C. \: \: \: \: F = qq/kr^ 2 \\D. \: \: \: \: F = kr^2 /qq \\ [/tex]

F is directly proportional to the product of the charges

[tex]F∝qq[/tex]

F is inversely proportional to the square of the distance between them

[tex]F∝ \frac{1}{ {r}^{2} } [/tex]

from above 2 equation:-

we get:-

[tex]F∝ \frac{qq}{ {r}^{2} } [/tex]

To remove proportionality sign we use constant for this case we r using constant k

[tex]F = \frac{Kqq}{ {r}^{2} } [/tex]

So your answer is :-

OPTION A.

[tex]F = \frac{Kqq}{ {r}^{2} } [/tex]

Answer:

• Potential Difference (V) = 2000 V

• Distance b/w the two parallel plates (d) = 10 cm = 10/100 = 1/10 = 0.1 m

• Electric field (E) = ?

[tex]\implies V = E.d[/tex]

[tex]\implies E = \dfrac{V}{d} [/tex]

[tex]\implies E = \dfrac{2000}{0.1} [/tex]

[tex]\implies E = \dfrac{2000}{ {10}^{ - 1} } [/tex]

[tex]\implies E = 2000 \times {10}^{1} [/tex]

[tex]\implies\bf E = 20000 \:V/m[/tex]

[tex]\implies\bf E = 20\:kV/m[/tex]

Hence, option C) the correct answer.

Answer:

a)  v = 0.18 m / s, b)  K₀ₐ = 0.0378 J, c) K_{ob}= 0, d)  K = 0.02349 J,

Explanation:

a) For this exercise we must define a system formed by the two cars, so that the forces during the collision are internal and the moment is conserved

initial instant. Before the hole

         p₀ = ma v₀ₐ

final intnate. After the crash

         p_f = (mₐ + m_b) v

the moment is preserved

         p₀ = p_f

         mₐ v₀ₐ = (mₐ + m_b) v

         v = [tex]\frac{m_a}{m_a+m_b} \ v_{oa}[/tex]

let's calculate

         v = [tex]\frac{0.900}{0.900+0.550} \ 0.29[/tex]

         v = 0.18 m / s

in the same direction of the movement of carriage A

b) the initial kinetic energy car A

         K₀ₐ = ½ m  v₀ₐ²

         K₀ₐ = ½ 0.900 0.29²

         K₀ₐ = 0.0378 J

c) kinetic energy of carriage B

          k_{ob} = 0

because the car is stopped

d) the kinetic energy of the system

          K = ½ (mₐ + m_b) v²

           K = ½ (0.900 + 0.550) 0.18²

           K = 0.02349 J

E) we see that part of the kinetic energy is lost, therefore the scientific reeling is not conserved in inelastic collisions

F) and momentum is conserved since it is equal to the variation of the moment and this is conserved in all collisions

Answer:

Train wheels and rails are both made of steel, and the steel-steel friction coefficient is around 0.25. As a result, the stopping time and distance will be three to four times that of a car.

Answer:

c). Easier setup

Explanation:

As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.

Answer:

the required solution is; x(t) = 0.675sin( 2.222t )

Explanation:

Given the data in the question;

Using both Newton's and Hooke's law;

m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0, [tex]x[/tex](0) = 0, [tex]x^f[/tex](0) = 1.5

given that mass m = 9 kg

[tex]x[/tex] = 1.8 m

k is F / x

hence

k = F / x

given that, F = 80 N

we substitute

k = 80 / 1.8

k = 44.44

so

m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0,

we input

9[tex]x^{ff[/tex] + 44.44[tex]x[/tex] = 0,

[tex]x^{ff[/tex] + 4.9377[tex]x[/tex] = 0

so auxiliary equation is,

r² + 4.9377 = 0

r² = -4.9377

r = √-4.9377

r = ±2.222i

hence, the solution will  be;

x(t) = A×cos( 2.222t ) + B×sin( 2.222t )

⇒ [tex]x^t[/tex](t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )

using initial conditions

x(0) = 0

⇒ 0 = A

[tex]x^t[/tex](t) = 1.5

1.5 = 2.222B

so

B = 1.5 / 2.222 = 0.675

Hence, the required solution is; x(t) = 0.675sin( 2.222t )

[tex]4.4×10^{-7}\:\text{m}[/tex]

Explanation:

The minimum energy needed to kick out an electron from a metal's surface is when the energy of the incident radiation is equal to the metal's work function [tex]\phi[/tex]:

[tex]E = h\nu - \phi = \dfrac{hc}{\lambda} - \phi = 0[/tex]

or

[tex]\dfrac{hc}{\lambda} = \phi[/tex]

Solving for the wavelength [tex]\lambda[/tex],

[tex]\lambda = \dfrac{hc}{\phi}[/tex]

[tex]\:\:\:\:\:=\dfrac{(6.62×10^{-34}\:\text{J-s})(3.0×10^8\:\text{m/s})}{4.5×10^{-19}\:\text{J}}[/tex]

[tex]\:\:\:\:\:= 4.4×10^{-7}\:\text{m}[/tex]

Note that as the radiation's wavelength increases, its energy decreases. So a radiation whose wavelength is longer than this maximum will lose its ability to kick out an electron from this metal.

The maximum wavelength, in nm, of light that can eject an electron from the metal, given the data is 441.73 nm.

To find the wavelength, the given values are,

Energy (E) = 4.50×10¯¹⁹ J

The distance between two consecutive crests and troughs is called the wavelength of a wave.

Here, for the wavelength,

Energy (E) = 4.50×10¯¹⁹ J

Planck's constant (h) = 6.626×10¯³⁴ Js

Speed of light (v) = 3×10⁸ m/s

The wavelength of the light can be obtained as illustrated below:

E = hv / λ

Cross multiply λ,

E × λ = hv

Divide both sides by E,

λ = hv / E

Substituting all the values,

λ = (6.626×10¯³⁴ × 3×10⁸) / 4.50×10¯¹⁹

λ = 0.000000441733 m

λ = 441.73nm

λ - The maximum wavelength of light.

Thus, the wavelength of the light that can eject an electron from the metal is 441.73 nm

Learn more about wavelength,

https://brainly.com/question/13047641

#SPJ2

Answer:

the person's heart rate

Answer:

Explanation:

- Pollution from cars

- Burning fossil fuels

Complete Question

The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20kg/m^2

Answer:

[tex]h=1614m[/tex]

Explanation:

From the question we are told that:

Initial Pressure [tex]P_1=980mbar=>98000Pa[/tex]

Final Pressure [tex]P_2=790mbar=>79000Pa[/tex]

Density [tex]\rho=1.20kg/m^2[/tex]

Generally the equation for Height climbed is mathematically given by

[tex]h=\frac{P_1-P_2}{\rho*g}[/tex]

[tex]h=\frac{P_1-P_2}{1.20*9.81}[/tex]

[tex]h=1614m[/tex]

Answer:

[tex]T=3*10^-3 N/m[/tex]

Explanation:

From the question we are told that:

Radius :

[tex]R_1=4=>0.04\\\\R_2=6=>0.06[/tex]

Work [tex]W=1.5 * 10^{-4}[/tex]

Generally the equation for Work done  is mathematically given by

[tex]W=TdA[/tex]

Where

[tex]dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)[/tex]

[tex]dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)[/tex]

[tex]dA=0.050m^2[/tex]

Therefore

[tex]W=TdA[/tex]

[tex]T=\frac{1.5 * 10^{-4}}{0.05m^2}[/tex]

[tex]T=3*10^-3 N/m[/tex]

Answer:

standard kilogram is the SI unit of mass

Answer:

The total mass of platinum-irridum cylinder whose diameter is equal to its height and stored at 0°C in the bureau of weight and measure in France is called 1 standard kilogram

Answer:

The initial angular speed of the CD is equal to 14.73 rad/s.

Explanation:

Given that,

Angular displacement, [tex]\theta=12\ rad[/tex]

Final angular speed, [tex]\omega_f=17\ rad/s[/tex]

The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]

We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,

[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]

Put all the values,

[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]

So, the initial angular speed of the CD is equal to 14.73 rad/s.

Answer:

If the person is to remain the floor the reaction force will be equal to the normal force exerted by the floor.

F(normal) - F(reaction) = 0

That means the person is not moving with respect to the elevator.

Expanding the applied forces we have:

Fw - Fn = 720 - 710 = 10 N   where the positive direction is chosen as down

Fw is the weight of the person and Fn the force exerted on the person by the elevator,

The acceleration of the person the becomes F = m a = m * 10 N and will be downward agreeing with our choice of coordinate axes.

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]

d = 68.5 x 10⁻⁶ m = 68.5 μm

Answer:

The impulse experienced by the object equals the change in momentum of the object. In equation form, F.t = m. Δv. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum.

Answer:

196 megajoules

Explanation:

Since you are talking about the gravitational potential I am assuming 10km is the height of the object in free fall.

PEg = mgh    2000kg×9.8m/s²×10000m = 196 megajoules

Answer:

answer : option (b) 0.016 amp

explanation : resistance of resistor , R = 10 Ω

inductance of inductor , X_LX

L

= 20H

voltage of AC circuit , V = 120volts

frequency, ff =60Hz

so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s

now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=

R

2

+ω

2

L

2

V

= 120/√{10² + (120π)² × 20²}

= 120/√{100 + 14400π² × 400}

after solving this we get, i = 0.016 amp

Answer:

Answer is B (Decomposition)

Sorry I really see ur questions but I don't know the answer but next time I will try to answer sorry:(