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2 years ago

Now, using your mass (in kg), and the figures for g (in the table below), you can calculate your weight on other planets.

Physics

1 answer:

Licemer1 [7]2 years ago

3 0

Answer:

1) Weight on Mercury

$F =W=mg=68.11 \times 3.61 m.s^{-2}$

Explanation:

do the same to the rest and use your calculator to find the weight in N.

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At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri

viva [34]

Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

Mass of the ball: $m_{1} =0.12kg$ ]

Initial velocity of the ball: $v_{i1}$

final velocity of the ball: $v_{f1}$ which is -30/100 of $v_{i1}$ = $-0.3v_{i1}$

Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

by using conservation momentum, which stated that the initial momentum is equal to the final momentum.

$m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$

so since the bottle is at rest firstly, therefore $v_{i2} =0$

$m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$

$m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ equation 1

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

$m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$

collect the like terms

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

6 0

2 years ago

Read 2 more answers

A manufacturer selected a metal to use in producing a lightweight button for clothing. A metal that has a density of 2.71 g/cm3

Natali5045456 [20]

Just find the density of every metal and select the one with a density of 2.71 g/cm³ . This is:

Metal 1

ρ = m/V

ρ = 22.1 g / 3 cm³

ρ = 7.367 g / cm³

Metal 2

ρ = m/V

ρ = 42 g / 4 cm³

ρ = 10.5 g / cm³

Metal 3

ρ = m/V

ρ = 9.32 g / 5 cm³

ρ = 1.864 g / cm³

Metal 4

ρ = m/V

ρ = 8.13 g / 3 cm³

ρ = 2.71 g / cm³

<h2>R / Metal 4 was selected.</h2>

4 0

2 years ago

What are the damages or effects of sewage dumps into natural water sources???

masha68 [24]

Answer:

Sewage dumps pollute our waterways causing death of marine life, water pollution, and disease.

Explanation:

Sewage dumps. bring disease or death to our ecosystems

5 0

2 years ago

Read 2 more answers

The magnetic domains in a non-magnetized piece of iron are characterized by which orientation

mel-nik [20]

The unmagnetized pieces of iron would be randomly pointing to directions, this is true because although influenced with the magnetic domain, the direction of the unmagnetized iron field of attraction is not uniform or does not have preferred direction.

3 0

3 years ago

Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. A particle beam is m

Genrish500 [490]

Answer:

the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

Explanation:

Given the data in the question;

Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J

Mass of proton = 1.673 × 10⁻²⁷ kg

Charge of proton = 1.602 × 10⁻¹⁹ C

distance d = 2 m

we know that

Kinetic Energy = Charge of proton × Potential difference ΔV

Potential difference ΔV = Kinetic Energy / Charge of proton

we substitute

Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )

Potential difference ΔV = 20287.14 V

Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;

E = Potential difference ΔV / distance d

we substitute

E = 20287.14 V / 2 m

E = 10143.57 V/m or 1.01 × 10⁴ V/m

Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

3 0

2 years ago