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2 years ago

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The weight (W) of an object is the product of its mass (m) in kilograms and g, the acceleration due to gravity in meters/second2

. If the acceleration due to gravity is 9.8 meters/second2, find the corresponding weights for these masses.
mass in kilograms = {22.1, 33.5, 41.3, 59.2, 78}

The weights (in newtons) for the given masses are

1 answer:

Nookie1986 [14]2 years ago

7 0

The weights in newtowns for the given masses are

<span> masses 22.1, 33.5, 41.3, 59.2, 78
weights 216.58N 328.3N 404.74N 580.16N 764.4N

e.g, for m=22.1kg, W=22.1kgx9.8N/kg =216.58N</span>

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If you wanna make it easier you can number them going down 1,2,3, or 4 & just tell me the number

prisoha [69]

the number is 1 I hope this help

8 0

2 years ago

Based on the Law of Conservation of Energy, which of the below is true?(1 point)

Veseljchak [2.6K]

Answer:

C

Explanation:

why because if something is conserved, it is constant, and does not change with time. A moving body may change its position, acceleration, and velocity with time, but it's energy is constant. The conversation of energy law states that: In any closed system (isolated system) the total energy of the system remain constant.

Mathematically it is written as

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8 0

1 year ago

A metallic conductor has a resistivity of 35 × 10 −6 Ω⋅m. What is the resistance of a piece that is 20 m long and has a uniform

Nitella [24]

Answer:

Therefore the resistance of the conductor is 175Ω

Explanation:

Resistance:

Resistance of a metallic conductor is directly proportional to its length(l).

Resistance of a metallic conductor is inversely proportional to its cross section area(A).

The notation sign of resistance is R.

The unit of resistance is ohm (Ω).

Therefore,

$R \propto l$

and

$R \propto \frac{1}{A}$

$\therefore R\propto \frac{l}{A}$

$\Rightarrow R=\rho \frac{l}{A}$

ρ is the proportional constant.

It is also known as resistivity of that metal.

Given ρ=35×10⁻⁶Ω-m

l= 20 m

A= 4.0×10⁻⁶m²

$\therefore R=35\times 10^{-6}\times \frac{20}{4.0\times 10^{-6}}$

=175Ω

Therefore the resistance of the conductor is 175Ω

3 0

2 years ago

a stone is thrown horizonttaly from a cliff of a hill with an initial velocity of 30m/s it hits the ground at a horizontal dista

ELEN [110]

Answer:

a) Time = 2.67 s

b) Height = 35.0 m

Explanation:

a) The time of flight can be found using the following equation:

$x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}$ (1)

Where:

$x_{f}$ : is the final position in the horizontal direction = 80 m

$x_{0}$ : is the initial position in the horizontal direction = 0

$v_{0_{x}}$ : is the initial velocity in the horizontal direction = 30 m/s

a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)

t: is the time =?

By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:

$t = \frac{x_{f}}{v_{0}} = \frac{80 m}{30 m/s} = 2.67 s$

b) The height of the hill is given by:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final position in the vertical direction = 0

$y_{0}$ : is the initial position in the vertical direction =?

$v_{0_{y}}$ : is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)

g: is the acceleration due to gravity = 9.81 m/s²

Hence, the height of the hill is:

$y_{0} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(2.67 s)^{2} = 35.0 m$

I hope it helps you!

5 0

2 years ago

What vertical distance Δy does a free-falling particle travel from the moment it starts to the moment it reaches a speed of 7.9

mr_godi [17]

Answer:

3.2 m

Explanation:

The equation to use to solve this problem is:

$v_f^2 = v_i^2 + 2 a \Delta y$

where

$v_f$ is the final velocity

$v_i$ is the initial velocity

a is the acceleration

$\Delta y$ is the distance covered

For the particle in free-fall in this problem, we have

$v_i = 0$ (it starts from rest)

$v_f = 7.9 m/s$

$g=9.8 m/s^2$ (acceleration due to gravity)

By re-arranging the equation, we can find the distance travelled:

$\Delta y = \frac{v_f^2 -v_i^2}{2a}=\frac{(7.9 m/s)^2-0^2}{2(9.8 m/s^2)}=3.2 m$

5 0

3 years ago