FASTTT FOR BRAINLIST

The ballistic pendulum is a device used to measure the speed of a projectile such as a bullet. The projectile of mass m is fired

My name is Ann [436]

Answer:

Relation between initial speed of bullet and height h is given as

$v = \frac{m + M}{m}\sqrt{2gh}$

Explanation:

As we know that system of block and bullet swings up to height h after collision

So we have

$(m + M)gh = \frac{1}{2}(m + M)v_1^2$

so we have

$v_1 = \sqrt{2gh}$

so speed of the block + bullet just after the impact is given by above equation

Now we also know that there is no force on the system of bullet + block in the direction of motion

So we can use momentum conservation

$mv = (m + M)v_1$

now we have

$v = \frac{m + M}{m}\sqrt{2gh}$

5 0

3 years ago

Stars are formed in which of the following?

MAXImum [283]

Stars are formed in nebulas, interstellar clouds of dust and gas.

6 0

3 years ago

After a hold up, Robin Banks flees in his 1575 kg getaway car at 20 m/s. He crashes into a 45 kg highway barrel which is at rest

fiasKO [112]

Answer:

v = 38.9 m/s

Explanation:

7 0

3 years ago

Read 2 more answers

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 
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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 energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard

gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

$|\Delta E_{ij}| = |E_i-E_j|$

Our states are given by

$E_1 = -11.7eV$

$E_2 = -4.2eV$

$E_3 = -3.3eV$

In this way the energy balance for the states would be given by,

$|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\$

$|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV$

$|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV$

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0

4 years ago