2. A 2500 kg car is slowed down uniformly from an initial velocity of 20.0 m/s to
the north by a 6250 N braking force acting opposite the car's motion. Use the
impulse-momentum theorem to answer the following questions:
a. What is the car's velocity after 2.50 s?
b. How far does the car move during 2.50 s?
c. How long does it take the car to come to a complete stop?

Answer:

it is A or D

Explanation:

Answer:

ANswer:A

Explanation:

Answer:

Inductive reactance is 50.00 ohms

Explanation:

Given the following data;

Voltage = 120v

Frequency = 60Hz

Current = 2.4 A

To find the inductive reactance;

Inductive reactance, XL = V/I

Where;

Substituting into the equation, we have;

XL = 120/2.4

XL = 50.00 ohms

Answer:

Average speed = 10,000 m/s

Explanation:

Given the following data;

Distance = 2m

Time = 0.0002secs

To find the average speed;

Average speed = distance/time

Average speed = 2/0.0002

Average speed = 10,000 m/s

Therefore, the average speed of the

electron is 10,000 meters per seconds.

Answer:

F = 482.51 N

Explanation:

Given that,

Mass of a child, m = 22 kg

Angular velocity of the merry-go-round, [tex]\omega=40\ rev/min[/tex]

Let the radius of the path, r = 1.25 m

We need to find the centripetal force acting on the child. The formula for the centripetal force is given by :

[tex]F=m\omega^2r\\\\=22\times (4.18879)^2\times 1.25\\\\=482.51\ N[/tex]

So, the required centripetal force is 482.51 N.

Answer:

1.84 m

Explanation:

For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.

So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,

a = g + rω²

= 9.8 m/s² + 1.10 m × (18.85 rad/s)²

= 9.8 m/s² + 390.85 m/s²

= 400.65 m/s²

Now, using v² = u² + 2a(h₂ - h₁)  where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.

v² = u² + 2a(h₂ - h₁)

So, v² - u² = 2a(h₂ - h₁)

h₂ - h₁ =  (v² - u²)/2a

h₂ =  h₁ + (v² - u²)/2a

substituting the values of the variables into the equation, we have

h₂ =  1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)

h₂ =  1.3 m + [-430.15 (m/s)²]/-801.3 m/s²

h₂ =  1.3 m + 0.54 m

h₂ =  1.84 m

Answer:

t = 0.196 s

Explanation:

The speed of a pulse is determined by the characteristics of the medium, its density and its resistance to stress, as long as these remain the speed will be constant for which we can use the kinetic expressions of the uniform movement

          v = x / t

          t = x / v

calculate

          t = 2/102

          t = 0.196 s

The question is incomplete, here is the complete question:

Which is greater, the energy of one photon of orange light or the energy of one quantum of radiation having a wavelength of [tex]3.36\times 10^{-9}m[/tex]

Answer: The energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.

Explanation:

To calculate the energy of one photon, we use the Planck's equation:

[tex]E=\frac{N_Ahc}{\lambda}[/tex]

where,

E = energy of radiation

[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}mol^{-1}[/tex]

h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]

c = speed of light = [tex]3\times 10^8 m/s[/tex]

[tex]\lambda}[/tex] = wavelength of radiation

For 1 photon, the term [tex]N_A[/tex] does not appear

[tex]\lambda}[/tex] = 620 nm = [tex]620\times 10^{-9}m[/tex]             (Conversion factor: [tex]1nm=10^{-9}m[/tex] )

Putting values in above equation, we get:

[tex]E=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{620\times 10^{-9}m}\\\\E=3.206\times 10^{-19}J[/tex]

[tex]\lambda}[/tex] = [tex]3.36\times 10^{-9}m[/tex]

Putting values in above equation, we get:

[tex]E=\frac{6.022\times 10^{23}mol^{-1}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.36\times 10^{-9}m}\\\\E=3.56\times 10^{7}J/mol[/tex]

Hence, the energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.

Answer:

0.1Ns

Explanation:

Impulse is the product of Force and time

Impulse = Force * Time

Given

Force = 10N

Time = 0.01s

Substitute into the formula

Impulse = 10 * 0.01

Impulse = 10 * 1/100

Impulse = 10/100

Impulse = 0.1Ns

hence the impulse of the hammer is 0.1Ns

Answer:

Explanation:

We shall apply Doppler's effect of sound .

speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at  6 m/s .

apparent frequency = [tex]f_o\times\frac{V+v_o}{ V-v_s}[/tex]

V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .

Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f

apparent frequency =  [tex]f\times \frac{340+6}{340-10}[/tex]

= [tex]f\times \frac{346}{330}[/tex]

So m = 346 , n = 330 .