Ask question

Ask question

All categories

3 years ago

6

Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.

3) 10 units 110 units 50 units 30 units

1 answer:

Yanka [14]3 years ago

4 0

Answer:

the correct answer is option C which is 50 units.

Explanation:

given,

two vector of magnitude = 30 units and of 70 units

to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}

cos θ value varies from -1 to 1

so, resultant vector

= $\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}$

a = 30 units and b = 70 units

= $\sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}$

= 40 units to 100 units

hence, the correct answer is option C which is 50 units.

You might be interested in

A geologic event causes changes to the physical makeup of a particular place and occurs _____.

babymother [125]

A geologic event causes changes to the physical makeup of a particular place and occurs slowly.

Geological events are what causes numerous changes and phenomena on the Earth's surface. Examples of these events include cliff erosion, volcanic eruption, or sedimentation at a mouth of a river.

Geological processes are extremely slow. However, because of the immense lengths of time involved, huge physical changes do occur - mountains are created and destroyed, continents form, break up and move over the surface of the Earth, coastlines change and rivers and glaciers erode huge valleys.

Geological events are both classified as internal and external. This means that these events occur both in the Earth's surface and interior.

4 0

3 years ago

Read 2 more answers

Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, by amplifying the input force

marusya05 [52]

Answer:

The solution and the explanation are in the Explanation section.

Explanation:

According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:

TA = W * (a * cosθ)

The torque due to effort at C point is:

TC = E * (b * cosθ)

The net torque is equal to 0, we have:

Tnet = 0

W * (a * cosθ) - E * (b * cosθ) = 0

$E=W\frac{a}{b}$

From the figure, you can observe that a/b < 1, thus E < W

8 0

3 years ago

What is Potentiometer

Katena32 [7]

Answer:

an instrument for measuring an electromotive force by balancing it against the potential difference produced by passing a known current through a known variable resistance.

3 0

2 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

2 years ago

A frictionless pendulum is made with a bob of mass 19.7 kg. The bob is held at height = 0.934 meter above the bottom of its traj

Alika [10]

Answer:

265 J

Explanation:

$Energy=PE+KE=mgh+ 0.5mv^{2}$ where KE is kinetic energy, PE is potential energy, m is the mass of an object, v is the speed, h is the height and g is acceleration due to gravity.

Substituting 19.7 Kg for mass, 0.934 for h, 2.93 for v and 9.81 for g then

$Energy=19.7(9.81*0.934+0.5*2.93^{2})=265.063303\approx 265 J$

4 0

2 years ago