What is a difference between in the field and in the laboratory

If a truck runs into a wall and stops, the truck loses momentum. Because

nignag [31]

The momentum goes to the wall

6 0

3 years ago

A train is accelerating at a rate of 2 km/hr/s. If its initial velocity is 20 km/hr, what is its velocity after 30 seconds?

Mademuasel [1]

Here it is. *WARNING* VERY LONG ANSWER

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11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h) 

The change in PE =mgh=5*9.8*12=588 J 
______________________________________... 
12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s. 

Commendable and jockey Pat Day had a combined mass =M= 550.0 kg, 

Their KE as they crossed the line=(1/2)Mv^2 

Their KE as they crossed the line=0.5*550*(15.98)^2 

Their KE as they crossed the line is 70224.11 J 

______________________________________... 
13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m 

She trips and drops the spare tire of mass = m = 10.0 kg, 

The tire rolls down the hill with an intial speed = u = 2.00 m/s. 

The height of top of the next hill = h = 5.00 m 

Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2 

Initial total mechanical energy =mgH+(1/2)mu^2 

Suppose the final speed at the top of second hill is v 

Final total mechanical energy =PE+KE=mgh+(1/2)mv^2 

As mechanical energy is conserved, 

Final total mechanical energy =Initial total mechanical energy 

mgh+(1/2)mv^2=mgH+(1/2)mu^2 

v = sq rt [u^2+2g(H-h)] 

v = sq rt [4+2*9.8(20-5)] 

v = sq rt 298 

v =17.2627 m/s 

The speed of the tire at the top of the next hill is 17.2627 m/s 
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14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean. 

a.)The mass of bean = m = 2.0 g 

Height up to which the been jumps = h = 1.0 cm from hand 

Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg 

b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s 
_____________________________ 
15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill. 

The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s 
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16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m, 

The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2 
______________________________________... 

EDIT 
1.) A train is accelerating at a rate = a = 2.0 km/hr/s. 

Acceleration 

Initial velocity = u = 20 km/hr, 

Velocity after 30 seconds = v = u + at 

Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s = 

Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr 

Velocity after 30 seconds = v = 80 km/hr=22.22 m/s 
_______________________________- 
2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins. 

His acceleration = a =11.1/9=1.233 m/s^2 

Distance he covered = s = (1/2)at^2=49.95 m

7 0

3 years ago

The half-life of plutonium 239 is 24,200 years. Assume that the decay rate is proportional to the amount. Determine the amount o

kotykmax [81]

Answer:

time taken is equal to 14,156 years

Explanation:

we know,

$Y=Ae^{-kt}$

at t = 0

Y(0) = A

given that half life of plutonium 239 = 24,200

$\dfrac{A}{2}=Ae^{-kt}\\0.5=e^{-kt}\\k\times 24200 = ln(2)\\k = \dfrac{ ln(2)}{24200}$

$Y=Ae^{-kt}$

$\frac{3}{2} = e^{-kt}\\ln(1.5)=-\dfrac{ ln(2)}{24200}\times t\\t=-\dfrac{ln(1.5)\times 24200}{ ln(2)}\\t=14,156 \ years$

hence time taken is equal to 14,156 years

5 0

3 years ago

Consult Multiple-Concept Example 5 to review the concepts on which this problem depends. A light bulb emits light uniformly in a

den301095 [7]

Answer:

a. 0.332 W/m² b. 11.2 V/m c. 15.8 V/m

Explanation:

(a) the average intensity of the light,

Intensity, I = P/A where P = average power = 150.0 W and A = area through which the power emits = 4πr² where r = distance from bulb = 6 m.

So, I = P/A = P/4πr²

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

I = P/4πr²

I = 150.0 W/4π(6 m)²

I = 150.0 W/4π(36 m²)

I = 150.0 W/452.39 m²

I = 0.332 W/m²

(b) the rms value of the electric field,

Since Intensity, I = E²/cμ₀ where E = rms value of electric field, c = speed of light = 3 × 10⁸ m/s and μ₀ = permeability of free space = 4π × 10⁻⁷ H/m.

Making E subject of the formula, we have

E² = Icμ₀

E = √(Icμ₀)

Since I = 0.332 W/m², substituting the other terms into the equation, we have

E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)

E = √(12.5 × 10)

E = √125 V/m

E = 11.18 V/m

E ≅ 11.2 V/m

(c) the peak value of the electric field.

The peak value of electric field, E' is gotten from E = E'/√2 where E = rms value of electric field.

So, E' = √2E

= √2 × 11.2 V/m

= 15.81 V/m

≅ 15.8 V/m

6 0

3 years ago

At highway speeds a car can accelerate at 1.7 m/s2.

Vaselesa [24]

Answer: 2.5 seconds

Explanation:

We know that the acceleration is:

a(t) = 1.7 m/s^2

To get the velocity function, we must integrate over time, and we will get:

v(t) = (1.7m/s^2)*t + v0

Where v0 is the initial velocity, in this case, we assume that we start at 23.6m/s, then the initial velocity is:

v0 = 23.6 m/s

Then the velocity equation is:

v(t) = (1.7m/s^2)*t + 23.6 m/s

Now we want to find the value of t such v(t) = 27.8 m/s

Then:

v(t) = 27.8 m/s = (1.7m/s^2)*t + 23.6 m/s

27.8 m/s - 23.6 m/s = (1.7m/s^2)*t

4.2 m/s = (1.7m/s^2)*t

4.2m/s/(1.7m/s^2) = t = 2.5 s

Then at that acceleration, you need 2.5 seconds.

6 0

2 years ago