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juin [17]
3 years ago
5

What is the final velocity of a dragster that travels 400 meters from the start line

Physics
1 answer:
krek1111 [17]3 years ago
8 0

Answer:

80m/s

Explanation:

The formula for velocity is [ v = s/t ] where s = displacement and t = time.

= 400/5

= 80m/s

Best of Luck!

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Daffy Duck is standing 6.8 m away from Minnie Duck. The attractive gravitational force between them is 5.4x10-8 N. If Daffy Duck
artcher [175]

Answer:

432.78 Kg

Explanation:

From the question given above, the following data were obtained:

Distance apart (r) = 6.8 m

Force of attraction (F) = 5.4×10¯⁸ N

Mass of Daffy Duck (M₁) = 86.5 kg

Mass of Minnie Duck (M₂) =?

NOTE: Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

The mass of Minnie Duck can be obtained as follow:

F = GM₁M₂ / r²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 6.8²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 46.24

Cross multiply

6.67×10¯¹¹ × 86.5 × M₂ =5.4×10¯⁸ × 46.24

Divide both side by 6.67×10¯¹¹ × 86.5

M₂ = 5.4×10¯⁸ × 46.24 / 6.67×10¯¹¹ × 86.5

M₂ = 432.78 Kg

Therefore, the mass of Minnie Duck is 432.78 Kg

8 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

________________________________________... 
<span>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) </span>

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

<span>Commendable and jockey Pat Day had a combined mass =M= 550.0 kg, </span>

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

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

<span>Their KE as they crossed the line is 70224.11 J </span>

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

<span>She trips and drops the spare tire of mass = m = 10.0 kg, </span>

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

<span>The height of top of the next hill = h = 5.00 m </span>

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

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

<span>Suppose the final speed at the top of second hill is v </span>

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

<span>As mechanical energy is conserved, </span>

<span>Final total mechanical energy =Initial total mechanical energy </span>

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

<span>v = sq rt [u^2+2g(H-h)] </span>

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

<span>v = sq rt 298 </span>

<span>v =17.2627 m/s </span>

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

<span>a.)The mass of bean = m = 2.0 g </span>

<span>Height up to which the been jumps = h = 1.0 cm from hand </span>

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

<span>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 </span>
<span>_____________________________ </span>
<span>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. </span>

<span>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 </span>
<span>__________________________________ </span>
<span>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, </span>

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

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

<span>Acceleration </span>

<span>Initial velocity = u = 20 km/hr, </span>

<span>Velocity after 30 seconds = v = u + at </span>

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

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

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

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

<span>Distance he covered = s = (1/2)at^2=49.95 m</span>
7 0
3 years ago
Please help on this one?
djyliett [7]
It’s C
Cause Impulse is found by multiplying the force and change in time (which is simply time)
So if you rearrange the equation for time you end up dividing Impulse by force.
3 0
3 years ago
What fraction of all the electrons in a 25 mg water
mihalych1998 [28]

Answer:

9.11\times 10^{-15}.

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of  -40 nC is removed from the water droplet.

The charge on one electron, \rm e=-1.6\times 10^{-19}\ C.

Let the N number of electrons have charge -40 nC, such that,

\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.  

Now, mass of one electron = \rm 9.11\times 10^{-31}\ kg.

Therefore, mass of N electrons = \rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is 25\ \rm mg = 25\times 10^{-6}\ kg.

Then,

\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.

It is the required fraction of mass of the droplet.

3 0
3 years ago
Loghan and Kyle want to see if the type of ice cream in the freezer affects how hard or soft it is. Each time they placed a new
poizon [28]
True.....................
7 0
3 years ago
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