3 years ago

1. Which of these is a relative adverb? where lovely who

Physics

2 answers:

PtichkaEL [24]3 years ago

3 0

Who is the right answer

Angelina_Jolie [31]3 years ago

3 0

the answer is where

Explanation:

none

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Using the knowledge you gained from your lessons and from this practical exercise, this engine appears to be a

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D)
(()(){({)({)(()({)())({)()({)()()()9

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3 years ago

Which of these quantities needs to have a direction associated with it? Select one: a. force only b. force and acceleration c. m

ehidna [41]

B) force & acceleration

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3 years ago

(b) The density of aluminum is 2.70 g/cm3. The thickness of a rectangular sheet of aluminum foil varies

vekshin1

Answer:

(i) $22.48 cm^3$

(ii) $1.5 mm$

Explanation:

Let t be the average thickness of the sheet.

Given that:

Density of the aluminum sheet is $2.70 g/cm^3$

Mass of sheet = 60.7 g

Length of sheet = 50.0 cm

Width of sheet = 30.0 cm

(i) Using, Density=Mass/Volume

$\Rightarrow \text{Volume}=\frac{\text{Mass}}{\text{Density}}$

$\Rightarrow \text{Volume}=\frac{60.7}{2.7}=22.48 cm^3$

Hence, the volume of the sheet is $22.48 cm^3$ .

(ii) Now, as this aluminum sheet is in the shape of a cuboid, so the volume of the sheet is

$\text{Volume}=\text{Length}\times\text{Width}\times\text{Thickness}$

$\Rightarrow 22.48=50\times 30 \times w$

$\Rightarrow w=\frac{22.48}{50\times 30}=0.015 cm$

Hence, the average thickness of the sheet is $1.5 mm$ .

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4 years ago

How much work is done if you push a box 200 meters with a force of 35 newtons answer?

VashaNatasha [74]

Work = force × distance
= 35 N × 200 m
= 7000 J

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3 years ago

A 1.6-kg ball is attached to the end of a 0.40-m string to form a pendulum. This pendulum is released from rest with the string

Digiron [165]

Answer:

a. 1.5 m/s

Explanation:

We will apply the law of conservation of energy in this situation between the initial position and the lowest point of the swing:

$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\\$

where,

m₁ = mass of ball = 1.6 kg

m₂= mass of block = 0.8 kg

u₁ = initial speed of ball = 0 m/s

u₂ = initial speed of block = 0 m/s

v₁ = final speed of ball = ?

v₂ = final speed of block = 3 m/s

Therefore,

$(1.6\ kg)(0\ m/s)+(0.8\ kg)(0\ m/s)=(1.6\ kg)(v_{1})+(0.8\ kg)(3\ m/s)\\\\v_{1} = \frac{2.4\ N.s}{1.6\ kg}\\$

v₁ = 1.5 m/s

Therefore, the correct option is:

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3 years ago