All categories

3 years ago

Vector B has x y, and z components of 8.8,

Physics

1 answer:

Ostrovityanka [42]3 years ago

3 0

Answer: 14.122 units

Explanation:

We have the following vector:

$\vec{V}=(x,y,z)=(8.8, 7.4, 8.2)$

If we want to find its magnitude we have to use the following formula:

$V=\sqrt{x^{2}+y^{2}+z^{2}}$

$V=\sqrt{(8.8units)^{2}+(7.4units)^{2}+(8.2 units)^{2}}$

$V=14.122 units$ This is the magnitude of vector $\vec{V}$

You might be interested in

Someone please help me

bagirrra123 [75]

Oxygen. Plants need water sunlight and CO2 to make their food. They get the carbon and the energy from the CO2 and the sun, and they need water as well. Then they release oxygen

3 0

3 years ago

Juan measured the temperature of salt water. He then added 273 to the measured value. Which conversation is Juan most likely doi

Vinvika [58]

The formula for Celsius to Kelvin is C+273=K

7 0

4 years ago

Take thepoints keep up the hard work

Fynjy0 [20]

Answer:

thanks

Explanation:

have a nice day

brainliest?

3 0

3 years ago

What do pressure and temperature determine about a gas

NikAS [45]

Answer:

With Boyle's law we have that for a constant temperature and gas quantity the pressure of a gas multiplied by its volume is also constant: This means that under the same temperature, two gases with equal quantity of molecules and equal volume must also have the same pressure, as well as that two gases with equal quantity and pressure must have the same volume.

Explanation:

8 0

3 years ago

As you travel from Detroit in a certain direction, the outside temperature, T (in degrees), depends on your distance, d (in mile

Ber [7]

Answer:

a) $\Delta T= 100^{\circ}C$

b) $\bigtriangledown T=1^{\circ}C.mile^{-1}$

c) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

d) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

Explanation:

Given is the data of variation of temperature with respect to the distance traveled:

Temperature T as a function of distance d:

$T=(d+30) ^{\circ}C$ ...................................(1)

(a)

Total change in temperature from the start till the end of the journey:

$\Delta T= T_f-T_i$ ..............................(2)

where:

$T_f$ = final temperature

$T_i$ = initial temperature

∵In the start of the journey d = 0 miles & at the end of the journey d = 100 miles.

So, correspondingly we have the eq. (2) & (1) as:

$\Delta T= (100+30)-(0+30)$

$\Delta T= 100^{\circ}C$

(b)

Now, the average rate of change of the temperature, with respect to distance, from the beginning of the trip to the end of the trip be calculated as:

$\bigtriangledown T=\frac{\Delta T}{\Delta d}$ ......................(3)

where:

$\Delta d$ = change in distance

$\bigtriangledown T=$ change in temperature with respect to distance

putting the respective values in eq. (3)

$\bigtriangledown T=\frac{100}{100}$

$\bigtriangledown T=1^{\circ}C.mile^{-1}$

(c)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

(d)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

4 0

4 years ago