Ask question

Ask question

All categories

2 years ago

5

The dimensions of a room are 16.40 m long, 4.5 m wide and 3.26 m high. What is the volume of the room in cubic meters? Express y

our answer in scientific notation.

1 answer:

anyanavicka [17]2 years ago

3 0

Answer:

the volume of the room is 240.588 meters3

The picture shows the scientific notation

You might be interested in

16. The energy absorbed in 10minutes by an electrical heater is 1.5 MJ. The supply voltage is 240 V. calculate: a) The current d

dsp73

Answer:

If by 1.5 MJ you mean 1.5E6 Joules then

W = P t = power X time

W / t = P power

P = 1.5E6 J / 600 sec = 2500 J / s

P = I V

a) I = 2500 J/s / (240 J/c) = 10.4 C / sec = 10.4 amps

b) Q = I t = 10.4 C / sec * 300 sec = 3120 Coulombs

c) E = P * t = 2500 J / sec * 100 hr * 3600 sec / hr = 9.0E8 Joules

5 0

2 years ago

A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo

DIA [1.3K]

Answer:

$74^{\circ} C$

Explanation:

We are given that

Mass of glass, $m=300 g$

$T_1=23^{\circ}$

Volume, $V=236cm^3$

Mass of water= $density\times volume=1\times 236=236 g$

Density of water= $1g/cm^3$

Temperature of hot water, $T=87^{\circ}$

Specific heat of glass, $C_g=0.2cal/g^{\circ}C$

Specific heat of water, $C_w=1 cal/g^{\circ}C$

$Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)$

$Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)$

$Q_{glass}+Q_{water}=0$

$300\times 0.2(T_f-23)+236\times 1(T_f-87)$

$60T_f-1380+236T_f-20532=0$

$296T_f=20532+1380=21912$

$T_f=\frac{21912}{296}=74^{\circ} C$

5 0

2 years ago

Please can anybody tell me what are these lab equipments

olga nikolaevna [1]

Answer:

5 is the tripoid stand

Thanks have a bangtastic day

4 0

2 years ago

Read 2 more answers

A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what

irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

$\Delta L=\frac{PL}{AE}$

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

$A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2$

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

$0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg$

Mass of the climber = 69.38 kg

6 0

3 years ago

What are the factors that effect the strength of a frictional force???

seropon [69]

If you write down the formula for friction, you will get an answer.

Ff = u * N Where N is a push down force that an object experiences.

u (mu) is a constant and has no units

It may not be accelerating and still experience friction. A is not correct.

Color and Density will not affect the frictional force. B is not so.

Buoyant forces are a different thing altogether. Generally friction has nothing to do with them. C is incorrect.

The last one is your answer. Technically mg should be the answer and not mass, but the second part is correct.

5 0

2 years ago