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bixtya [17]
3 years ago
10

What is the relationship between mass and weight

Physics
2 answers:
maxonik [38]3 years ago
6 0
Mass is how something moves and weight is how heavy something is
sergiy2304 [10]3 years ago
5 0
Mass is how heavy is it, weight is the size both are the same
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Two blocks are on a horizontal frictionless surface. Block a is moving with an initial velocity
AfilCa [17]

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

5 0
2 years ago
What is the meaning of the safety symbol shown below?
lora16 [44]

Answer:

D. Poison

Explanation:

8 0
2 years ago
Read 2 more answers
PLEASEEEE HELPPPPPPPPPPPP
lesantik [10]

Answer:

240 meter

Explanation:

d= s*d=60m/s * 4sec=240 meter

7 0
2 years ago
what is the resistance of a light bulb if a potential difference of 120 v will produce a current of 0.5 a in the bulb? 0.0042 0.
svlad2 [7]
R = V/I

R = 120 V/ 0.5 A

R = 240 ohms
3 0
3 years ago
Read 2 more answers
A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori
olasank [31]

Explanation:

The given data is as follows.

        k = 130 N/m,       \Delta x = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = \frac{1}{2}kx^{2}

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}

   \frac{1}{2} \times 130 \times (0.17)^{2} = \frac{1}{2} \times 2.8 \times v^{2}

          v^{2} = \frac{3.757}{2.8}

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = \mu \times mg

                                    = 0.15 \times 2.8 \times 9.8

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           \frac{1}{2}mv^{2} = F_{f} \times d

           \frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d

             1.8515 = 4.116 \times d

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0
3 years ago
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