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UNO [17]
3 years ago
13

A 2 kg blue car is moving 6 m/s to the right and collides with a 3 kg red car that is moving 2 m/s to the left. The cars collide

and stick together. What is their velocity after the collision? Include units on your answer.
Physics
1 answer:
snow_lady [41]3 years ago
7 0

Answer:

Their velocity after the collision is 1.2 m/s, to the right.

Explanation:

Given;

mass of the blue car, m₁ = 2 kg

initial velocity of the blue car, u₁ = 6 m/s

mass of the red car, m₂ = 3 kg

initial velocity of the red car, u₂ = 2 m/s

let the blue car moving to the right be in positive direction

also, let the red car moving to the left be in negative direction

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ - m₂u₂ = v(m₁ + m₂)

where;

v is their velocity after the collision

(2 x 6) - (3 x 2) = v(2 + 3)

12 - 6 = 5v

6 = 5v

v = 6/5

v = 1.2 m/s, to the right

Therefore, their velocity after the collision is 1.2 m/s, to the right.

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A skateboarder traveling with an initial velocity 9.0 meters per second,
meriva

Answer:

25m/s

Steps:

<em> First, The equation v= u + a * t shows us what we need to find, (the finale velocity). </em>

<em />

Second, we substitute the values given:

v= 9m/s + 4m/s2 * 4s

Last, We calculate the values:

Multiply 4m/s2 * 4s = 16m/s  

Add 9m/s + 16m/s

<u></u>

<u>Answer:  25m/s</u>

Hope this helps :)

4 0
3 years ago
g Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the bu
dlinn [17]

Answer:

the observed frequency will reduce but the wavelength will increase

Explanation:

As we know

fo = fs (v/(v-vs))

fo = observed frequency

vs = velocity of source

As per this equation,  

When an observer moves away from the stationary source, the observed frequency reduces. Since the observer in the balloon is moving away from the source which itself is moving in opposite direction, the observed frequency will reduce.  

Since wavelength = V/fs . The source frequency is unchanged but the velocity is increasing as it is moving in downward direction. Hence, the wavelength will increase

8 0
3 years ago
I need to choose a theme for my physics assignment My experiment is finding g
Kobotan [32]
<h3>Question:</h3>

How to find g (acceleration due to gravity)

<h3>Solution:</h3>

We know,

Acceleration due to gravity (g)

=  \frac{GM}{ {R}^{2} }

where, G = Gravitational constant

= 6.67 \times  {10}^{11} N {m}^{2}/k {g}^{2}  \\

M = Mass of the earth

= 6 \times  {10}^{24} \:  kg

R = Radius of the earth

= 6.4 \times  {10}^{6} m

Putting these values of G, M and R in the above formula, we get

g \:  =  \:  \frac{6.67 \times  {10}^{11} N {m}^{2}/k {g}^{2}   \times \: 6 \times  {10}^{24} \:  kg }{(6.4 \times  {10}^{6}m {)}^{2}  }  \\  = 9.8m/ {s}^{2}

So, the value of acceleration due to gravity is

9.8m/s ^{2}

Hope it helps.

Do comment if you have any query.

5 0
3 years ago
A 30 cm scale has one end broken. The mark at the broken end is 2.6 cm. How would you use it to measure the length of your penci
lidiya [134]

Answer:

the mark of the broken end is 2.6 cm so, we use the scale from the next full mark i.e. 3cm

Explanation:

<em>we </em><em>now </em><em>measure</em><em> </em><em>the </em><em>length</em><em> </em><em>of </em><em>the </em><em>pencil</em><em> </em><em>by </em><em>keeping </em><em>the </em><em>3</em><em> </em><em>c</em><em>m</em><em> </em><em>mark </em><em>of </em><em>the </em><em>scale</em><em> </em><em>at </em><em>it's</em><em> </em><em>left </em><em>end.</em>

<em>The </em><em>3</em><em> </em><em>cm </em><em>value </em><em>is </em><em>then </em><em>subtracted</em><em> </em><em>from </em><em>the </em><em>scale</em><em> </em><em>reading</em><em> </em><em>at </em><em>the </em><em>right</em><em> </em><em>side </em><em>end </em><em>of </em><em>the </em><em>pencil</em><em> </em><em>to </em><em>obtain </em><em>the </em><em>correct</em><em> </em><em>length</em><em> </em><em>of </em><em>the </em><em>pencil.</em><em> </em><em>✏️</em>

<em>(</em><em>i </em><em>i </em><em>)</em><em> </em>place the scale in the contact with object along it's length

(2) Your eyes must be exactly in front of the point where the measurements to be taken.

Hope_it_helps_mga_ka_joiners_mwehehe

6 0
3 years ago
A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
I= \frac{Q}{\Delta t}
First we need to find the net charge flowing at a certain point of the wire in one second, \Delta t=1.0 s. Using I=0.92 A and re-arranging the previous equation, we find
Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C

Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}
3 0
3 years ago
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