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IRINA_888 [86]
3 years ago
6

Newton’s first law relates motion to balanced and unbalanced forces.

Physics
2 answers:
BlackZzzverrR [31]3 years ago
8 0
True, an object at rest stays and rest and an object in motion stays in motion
gulaghasi [49]3 years ago
8 0

YES !  

True and certainly.  

Uh huh, uh huh.

That's a fact.  

More intense veracity is seldom encountered.

Did you have a question to ask ?

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In an elastic collision, _______ energy is conserved.
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Kinetic energy and momentum are conserved.

Explanation:

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Lithium (chemical symbol Li) is located in Group 1, Period 2. Which is lithium most likely to be? O A. A soft, shiny, highly rea
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A. A soft, shiny, highly reactive metal
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How are heat and light waves produced on and in the sun?
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All of the electromagnetic energy radiated from the sun (and from
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65. A length of wire is bent into a closed loop and a magnet is plunged into it, inducing a voltage and, consequently, a current
natulia [17]

Answer:

Resistance of the second wire is twice the first wire.

Explanation:

Let us first see the formula of resistance;

R = pxL/A

Here L is the lenght of the wire, A the area and p is the resistivity of wire.

As we are given that the length of second wire is double than that of the first wire, hence the resistance of second wire would be double.

Since we have two loop in second case, inducing double voltage but as resistance is doubled so the current would remain same according to ohms law

I = V/R

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3 years ago
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A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w
kramer

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = ?

v_{1i} = initial velocity of the first body before collision = v

v_{2i} = initial velocity of the second body before collision = 0 m/s

v_{1f} = final velocity of the first body after collision =

using conservation of momentum equation

m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}

Using conservation of kinetic energy

m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35

b)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = 1.35 kg

v_{1i} = initial velocity of the first body before collision = 4 ms⁻¹

v_{2i} = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67 ms⁻¹

8 0
2 years ago
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