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Inga [223]
2 years ago
11

I need help in physics, pls hurry! I will give Brainliest for correct answers!!

Physics
1 answer:
Nana76 [90]2 years ago
6 0
Top left: slowing down
Top right: not moving
Bottom left: moving at a constant speed
Bottom right: speeding up
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A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere
dmitriy555 [2]

Question: A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere, What is thier velocities after collision.

Answer:

v = 6 m/s, v' = 0 m/s

Explanation:

From the question,

For Elastic collision,

mu+m'u' = mv+m'v'......................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, v = final veolocity of the first sphere, v' = final velocity of the second sphere.

Also,

The relative velocity before collision = relative velocity after collision

u-u' = v-v'............................ Equation 2

Given:  m = 2 kg, m' = 10 kg, u = 6 m/s, u' = 0 m/s

Substitute into equation 1 and 2

2(6)+10(0) = 2v+10v'

2v+10v' = 12.............. Equation 3

6-0 = v-v'

v-v' = 6 ................... Equation 4

Solve equation 3 and 4 simultaneously.

v = 6+v'............. Equation 5

Substitute equation 5 into equation 3

2(6+v')+10v' = 12

12+2v'+10v' = 12

12v' = 12-12

v' = 0/12

v' = 0 m/s.

Also substitute the value of v' into equation 5

v = 6+0

v = 6 m/s

5 0
3 years ago
Physical properties of minerals graphic organizer
Nadusha1986 [10]
The answer is in the attachment
<span>...........................................</span>

4 0
3 years ago
An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu
ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, qE \times S + mg \times S = 0.5 \times mv^{2}

     1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

   (1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2&#10;}

         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

           = 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0
2 years ago
Cual sera el caudal que lleva un rio cuando se desplaza a 200 litros cada 40 segundos
alisha [4.7K]

Answer:

Q = 5 L/s

Explanation:

To find the flow you use the following formula (para calcular el caudal usted utiliza la siguiente formula):

Q=\frac{V}{t}

V: Volume (volumen) = 200L

t: time (tiempo) = 40 s

you replace the values of the parameters to calculate Q (usted reemplaza los valores de los parámteros V y t para calcular el caudal):

Q=\frac{200L}{40s}=5\frac{L}{s}

Hence, the flow is 5 L/s (por lo tanto, el caudal es de 5L/s)

4 0
3 years ago
The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions. On average, the river has
OverLord2011 [107]

Answer:

(a) V = 0.75 m/s

(b) V = 0.125 m/s

Explanation:

The speed of the flow of the river can be given by following formula:

V = Q/A

V = Q/w d

where,

V = Speed of Flow of River

Q = Volume Flow Rate of River

w = width of river

d = depth of river

A = Area of Cross-Section of River = w d

(a)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 20 m

d = 20 m

Therefore,

V = (300 m³/s)/(20 m)(20 m)

<u>V = 0.75 m/s</u>

<u></u>

(b)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 60 m

d = 40 m

Therefore,

V = (300 m³/s)/(60 m)(40 m)

<u>V = 0.125 m/s</u>

8 0
3 years ago
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