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3 years ago

What is the weight of a 5.00 kg object on Earth? Assume g=9.81 m/s^2.

Physics

1 answer:

Softa [21]3 years ago

7 0

weight = (mass) x (gravity)

Weight = (5.00 kg) x (9.81 m/s²)

weight = (5.00 x 9.81) (kg-m/s²)

Weight = 49.05 Newton

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A gas, behaving ideally, fills a fixed volume container at a pressure P₁ and at a temperature T₁. The temperature of the contain

Lynna [10]

Answer:

c. P₁/T₁=P₂/T₂

Explanation:

neither Avogadro’s, Charles’, or Boyle’s law formula can be used, since some parameters like volume is not given,

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3 years ago

A 250 g block of ice is removed from the refrigerator at -8.0°C. How much thermal energy does the ice absorb as it warms to room

zlopas [31]

Answer:

Q = 114895 J

Explanation:

To find the thermal energy gained by the ice you use the following formula:

$Q=mc(T_2-T_1)+H_f\ m$

m: mass of the ice = 0.250kg

T2: final temperature = 22°C

T1: initial temperature = -8.0°C

Hf: heat of fusion of water = 3.34*10^5 J/kg

c: specific heat of water = 4186 J/kg

By replacing the values of the parameters you have:

$Q=(0.250kg)(4186J/kg\°C)(22+8)\°C+(3.34*10^5 J/kg)(0.250kg)\\\\Q=114895\ J$

where you have considered that ice melts completely

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3 years ago

A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in air. A woman,

lina2011 [118]

Answer:

44.1 m

Explanation:

Given:

$V_a$ = speed of sound in air = 343 m/s
$V_r$ = speed of sound in the rod = $15V_a$
$\Delta t$ = times interval between the hearing the sound twice = 0.12 s

Assumptions:

$l$ = length of the rod
$t$ = time taken by the sound to travel through the rod
$T$ = time taken by the sound to travel to through air to the same point = $t+\Delta t = t+0.12\ s$

We know that the distance traveled by the sound in a particular medium is equal to the product of the speed of sound in that medium and the time taken.

For traveling sound through the rod, we have

$l=V_r t\\\Rightarrow t = \dfrac{l}{V_r}$ ..........eqn(1)

For traveling sound through the air to the women ear for traveling the same distance, we have

$l=V_aT\\\Rightarrow l=V_a(t+0.12)\\\Rightarrow l=V_a(\dfrac{l}{V_r}+0.12)\,\,\,\,\,\,(\textrm{From eqn (1)})\\\Rightarrow l=V_a(\dfrac{l}{15V_a}+0.12)\\\Rightarrow l=\dfrac{l}{15}+0.12V_a\\\Rightarrow l-\dfrac{l}{15}=0.12V_a\\\Rightarrow \dfrac{14l}{15}=0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = 44.1\ m$