What are scientists who study the forces and shape of the earth called?

Delvig [45]

Answer:

Approximately $73\; {\rm N}$ , assuming that the acceleration of this ball is constant during the descent.

Explanation:

Assume that the acceleration of this ball, $a$ , is constant during the entire descent.

Let $x$ denote the displacement of this ball and let $t$ denote the duration of the descent. The SUVAT equation $x = (1/2)\, a\, t^{2}$ would apply.

Rearrange this equation to find an expression for the acceleration, $a$ , of this ball:

$\begin{aligned} a &= \frac{2\, x}{t^{2}}\end{aligned}$ .

Note that $x = 11\; {\rm m}$ and $t = 1.5\; {\rm s}$ in this question. Thus:

$\begin{aligned} a &= \frac{2\, x}{t^{2}} \\ &= \frac{2 \times 11\; {\rm m}}{(1.5\; {\rm s})^{2}} \\ &\approx 9.78\; {\rm m \cdot s^{-2}}\end{aligned}$ .

Let $m$ denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is $a$ , the net external force on this ball would be $m\, a$ .

Since $m = 7.5\; {\rm kg}$ and $a \approx 9.78\; {\rm m\cdot s^{-2}}$ , the net external force on this ball would be:

$\begin{aligned} (\text{net force}) &= m\, a \\ &\approx 7.5\; {\rm kg} \times 9.78\; {\rm m\cdot s^{-2}} \\ &\approx 73\; {\rm kg \cdot m \cdot s^{-2} \\ &= 73\; {\rm N} && (1\; {\rm N} = 1\; {\rm kg \cdot m\cdot s^{-2}}) \end{aligned}$ .

4 0

3 years ago

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svlad2 [7]

Burning of paper is the answer

6 0

3 years ago

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It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J of energy whe

boyakko [2]

Answer:

10.4L of water is expended when 1L of crude oil is burned.

Explanation:

This problem requires us to calculate the volume of water that must be expended to absorb the amount of energy released from the burning of 1.00L of crude oil.

In going from water at 18.5°C to steam at 285°C, the water undergoes 3 stages:

Absorption of heat from 18.5°C to 100°C >> phase change to vapour at 100°C >> heating from 100°C to steam at 285°C

Given Cw = 4186 J/kg°C and Cs = 2020 J/kg°C. The latent heat of of vaporization of water Lv = 2.256 ×10⁶ J/kg

The total heat absorbed in the process per kilogram

H = Cw × (100 – 18.5) + Lv + Cs × (285 – 100)

H = 4186 × 81.5 + 2. 256 ×10⁶ + 2020× 185

= 2.696 ×10⁶ J/kg

The amount of heat in J/L of water needed can be calculated as follows:

Density of water = 1000kg/m³ =

1000 kg/m³ × 1m³/1000L = 1kg/L (Basically conversion of density in kg/m³ to kg/L)

Let the volume of water needed be V litres.

Then the mass of water that must be expended = Density × Volume