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lapo4ka [179]
3 years ago
5

Two blocks are on a horizontal frictionless surface. Block a is moving with an initial velocity

Physics
1 answer:
AfilCa [17]3 years ago
5 0

The final velocity of the block A will be 2.5 m/sec. The principal of the momentum conversation is used in the given problem.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

In a given concern, mass m₁ is M, mass m₂ is 3M. Initial speed for the mass m₁ and  m₂ will be u₁=5 and u₂=0 m/s respectively,

According to the law of conservation of momentum

Momentum before collision =Momentum after collision

m₁u₁+m₂u₂=(m₁+m₂)v

M×5+3M×0=[M+3M]v

The final velocity is found as;

V=51.25 m/s

The velocity of block A is found as;

\rm V_f=\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2} \\\\  V_f=\frac{(M-3M)5+2\times3M\times0}{m_1+m_2} \\\\ V_f=2.5 m/s

Hence, the final velocity of the block A will be 2.5 m/sec.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ4

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(a) Circular metal wires in electrical circuits may have different cross-sectional areas (different diameters) and different len
WARRIOR [948]

For the different wires the values given are

For Wire A,  Area=A_{A}, Resistivity=ρ_{A}, Length= L_{A}, Heat=H_{A}, Voltage= V, Time= t, Wire B,  Area=A_{B}, Resistivity=ρ_{B}, Length= L_{B}, Heat=H_{B}

So, H_{A}/ H_{A}=\frac{V^{2} *t}{\frac{ρ(1)*L_{A}  }{A_{A} } }/\frac{V^{2} *t}{\frac{ρ(2)*L_{B}  }{A_{B} } }

Therefore H_{A}/ H_{A}= \frac{A_{A}*ρ_{B} *L_{B}}{A_{B}ρ_{A} L_{A}  }

<h3>Heat</h3>

The type of energy that moves between two materials with varying temperatures is referred to by scientists as heat. Because the average translational kinetic energy per molecule in the two materials varies, an energy transfer takes place. Up until thermal equilibrium is attained, heat is transferred from the substance with the higher temperature to the material with the lower temperature. The joule, with 1 joule equalling 1 newton meter, is the SI unit of heat. Imagine the following situation to better comprehend what happens when this energy transfer takes place: Tiny rubber balls are bouncing all over two distinct containers that are full with them. The difference between the average ball speed in one container and the second container is substantial.

(a) Circular metal wires in electrical circuits may have different cross-sectional areas (different diameters) and different lengths. For a given applied voltage, how would the joule heat vary with these parameters

Learn more about heat here:

brainly.com/question/21041726

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6 0
2 years ago
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast wil
viktelen [127]

Answer:

a

The speed of the quarterback backward is v_q =  0.08 \ m/s

b

Known are

 m_Q , m_B , (v_{Bx})_i  (v_{Qx})_f, (v_{Bx})_f

Unknown

   (v_{Qx})_f

Explanation:

From the question we are told that

   The mass of the quarterback is m_Q =  80 \ kg

    The mass of the ball is m_B =  0.43 \ kg

     The speed of the ball is  v_{B x}=  15 \ m/s

The law of momentum conservation can be mathematically represented as

       m_Q u_{Qx} + m_Bu_{Bx}  =  - m_{Q} v_{Qx} + m_B v_{Bx}

Now at initial both ball and quarterback are at rest and the negative sign signify that the quarterback moved backwards after throwing the ball

  So

       m_Q v_{Qx} =  m_B v_ {Bx}

=>     v_{Qx} =  \frac{m_Bv_{Bx}}{m_Q}

substituting values

        v_q =  \frac{0.43 * 15}{80}

       v_q =  0.08 \ m/s

5 0
3 years ago
The specific heat of silver is 0.235 j/
zavuch27 [327]
The energy needed the raise the temperature of the block of silver is given by:
Q=mC_s \Delta T
where m is the mass of the block, Cs the specific heat capacity of silver and \Delta T its increase in temperature.

Using the numbers of the problem, we find
Q=(8 g)(0.235 J/gC)(50 C)=94 J
6 0
3 years ago
SOMEONE PLEASEEEEEEEEEEEEEEEEEEEE HELPPPPPPPPPPPPPPP
Evgen [1.6K]

Answer:

53.33 seconds

Explanation:

From the question;

  • Power of the motor is 75 kW or 75000 W
  • Depth or height is 150 m
  • Volume of water is 400 m³

We are required to determine taken to raise the water from the given height.

We know that density of water is 1000 kg/m³

Therefore;

Mass of water = 400 m³ × 1000 kg/m³

                       = 4.0 × 10^5 kg

Thus, force required to raise the water;

 = 4.0 × 10^5 kg × 10 N/kg

  = 4.0 × 10^6 N

To determine the time;

we use the formula;

 Time = work done ÷ power  

          = (4.0 × 10^6 N × 150 m) ÷ 75000 Joules/s

          = 53.33 seconds

Therefore, time taken to raise the water is 53.33 seconds

8 0
3 years ago
ANSWER BELOW QUESTIONS:
Bogdan [553]

Answer:

why should we do , do by your own , no sense

Explanation:

3 0
3 years ago
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