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Alla [95]
3 years ago
10

You drive a car 2.0 km to the gas station, 4.0 km to the shopping mall to let your mother off, and 4.5 km back home. The trip ta

kes 20.0 min. What is your average speed?
Physics
1 answer:
slamgirl [31]3 years ago
8 0
60 i believe  (ssssppppaaaccceeee)
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How many atoms are in a sample of 2.39 moles of neon (Ne) atoms?
Reil [10]
<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles.

 </span>2.39 moles Ne ( 6.022 x 10^23 atoms / mole ) = 1.44 × 10^24 atoms Ne
4 0
3 years ago
What is the force needed to move an 225 kg car a distance of 150 meters
Kobotan [32]

Physics

mass = 225 kg

weight = m × a = 2250 N

distance = 150 m

force : ______?

Answers :

W = F × s

W = 2250 × 150

W = 337500 ✅

4 0
3 years ago
Read 2 more answers
Daily life examples of pressure of solids on solids?​
Ipatiy [6.2K]

Explanation:

There's a massive amount, just think of anything everyday. Like a table on the floor, or when your walking around and putting pressure on the floor. When you squeeze something which is solid. Anything like that will do.

6 0
3 years ago
A11) A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic fiel
vitfil [10]

Answer:

A

Explanation:

From a Solenoid we know that a magnetic fiel is always inversely proportional to lenght L or BL = constant

B= frac{\mu_0}{2R}

As I is constant

\frac{B2}{B1} = \frac{R1}{R2}

B2 = 2mT*\frac{18}{21}

B2 = 1.714mT

7 0
3 years ago
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
3 years ago
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