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2 years ago

Identify the energy transformation in the image

Physics

1 answer:

Yuliya22 [10]2 years ago

4 0

Answer:

d one is correct

Explanation:

as the electrical energy in the socket is transferred to the electric tea pot

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Please help please help

Inessa [10]

Answer: p= m/v so 90kg/.075m^3 = 1,200

2a. .35 m 1.1 m and .015 m

2b. 35 cm x 110 cm x 1.5 cm = 5,775 cm^3 = 57.75 m^3

mass= pv

2700•57.75= 155,925 kg

mass= 155,925 kg

volume= 57.75 m^3

Explanation: physics

6 0

3 years ago

A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One

Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

$\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\$

To get the required work done, we will divide the mass by the speed of one knot to have:

$w=\frac{7230}{0.51}\\w= 14,176.47Joules$

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

Learn more here: brainly.com/question/25573786

8 0

2 years ago

1. Which of these is not a natural fibre?<br>a. leather<br>b. jute<br>C.Wool<br>d. cotton

Oksanka [162]

Answer:

leather

Explanation:

plz mark as brainliest......hope it helps

6 0

3 years ago

Read 2 more answers

65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight

Anna35 [415]

Answer:

Given, Apparent weight(W₂)=4.2N

Weight of liquid displaced (u)=2.5N

Let weight of body in air = W₁

Solution,

U=W₁-W₂

W₁=4.2=2.5=6.7N

∴Weight of body in air is 6.7N

5 0

2 years ago

A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?

anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}$

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

$\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}$

now, replace

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}$

hence, the answer is 0.495 m/s

3 0

1 year ago