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Triss [41]
3 years ago
5

Block A of 5 kg with a speed of 3 m/s collides with block B of 10 kg that is stationary. After the collision, block B travels wi

th a speed of 2 m/s
Physics
1 answer:
Lemur [1.5K]3 years ago
6 0

Answer:

-1m/s

Explanation:

We can calculate the speed of block A after collision

According to collision theory:

MaVa+MbVb = MaVa+MbVb (after collision)

Substitute the given values

5(3)+10(0) = 5Va+10(2)

15+0 = 5Va + 20

5Va = 15-20

5Va = -5

Va = -5/5

Va = -1m/s

Hence the velocity of ball A after collision is -1m/s

Note that the velocity of block B is zero before collision since it is stationary

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Which phrases describe all the outer planets’ motion? Check all that apply.
uranmaximum [27]

Answer: slow revolution and fast rotation

Solar system has 8 planets. 4 inner rocky planets - Mercury, Venus, Earth and Mars and 4 outer gaseous planets - <u>Jupiter, Saturn, Uranus and Neptune.</u>  The outer planets have few common features.

They are gaseous. There  period of revolution is larger than the inner planets which means that they have slow revolution about the Sun. One day on the outer planets is smaller than the inner planets which means they have fast rotation.

<u>For example,</u> Jupiter has revolves around sun in 11.86 Earth years and rotates about axis in 9.8 Earth hours. Uranus revolves around sun in 84 Earth years and rotates on its axis 17.9 Earth hours.

8 0
3 years ago
Read 2 more answers
A moonshiner makes the error of filling a glass jar to the brim and capping it tightly. The moonshine expands more than the glas
ohaa [14]

Answer:

ΔP = (640 N/cm^2)

Explanation:

Given:-

- The volume increase, ΔV/V0 = 4 ✕ 10^-3

- The Bulk Modulus, B = 1.6*10^9 N/m^2

Find:-

Calculate the force exerted by the moonshine per square centimeter

Solution:-

- The bulk modulus B of a material is dependent on change in pressure or Force per unit area and change in volume by the following relationship.

                                  B = ΔP / [(ΔV/V)]

- Now rearrange the above relation and solve for ΔP or force per unit area.

                                  ΔP = B* [(ΔV/V)]

- Plug in the values:

                                  ΔP = (1.6*10^9)*(4 ✕ 10^-3)

                                  ΔP = 6400000 N/m^2

- For unit conversion from N/m^2 to N/cm^2 we have:

                                  ΔP = (6400000 N/m^2) cm^2 / (100)^2 m^2

                                  ΔP = (640 N/cm^2)

7 0
3 years ago
A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
If the bond enthalpy for a C-H bond is 413 kJ, what will happen when the C-H bond is broken?
pentagon [3]

Explanation:

Bond Enthalpy : It is defined as amount of energy required to break a the particular bond in there gaseous state. It is also known as bond energy. It units are kJ/mol.

  • Breaking of a bond is an Endothermic process (energy absorbed from the surroundings).
  • Formation of bond is an Exothermic process (energy is released to the surroundings).

If the average bond enthalpy for a C-H bond is 413 kJ/mol, When the C-H bond breaks in which energy will be required ,which will be an endothermic reaction.

8 0
3 years ago
Juan is making ice tea. When he adds ice to the tea, why does the tea cool down?​
Tresset [83]

Answer: because the ice was cold, and it melted in the tea. So the tea is now cooler.

Explanation:

4 0
3 years ago
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