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astraxan [27]
3 years ago
11

C. 400 km/min

Physics
1 answer:
Nataly_w [17]3 years ago
5 0
A is the answer i hope
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The free-body diagram of a crate is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labe
satela [25.4K]

The net force acting on the crate is determined as 176 N to the left.

<h3>Net force acting on the crate</h3>

The net force acting on the crate is calculated as follows;

∑F = F1 + F2 + F3 + F4

F(net) = -440y + 176x + 440y - 352x

F(net) = -176 x

The resultant force is pointing in negative x direction.

Thus, the net force acting on the crate is determined as 176 N to the left.

Learn more about net force here: brainly.com/question/14361879

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8 0
1 year ago
A 388 Hz tuning fork is resonating in a closed tube on a warm day when the speed of sound is 346 m/s. What is the length of the
Marizza181 [45]

Answer:

A

Explanation:

because u are subtracting if this is from flvs that is what i did and it was right

5 0
3 years ago
Read 2 more answers
A carmaker has designed a car that can reach a maximum acceleration of 12 meters/second2. The car’s mass is 1,515 kilograms. Ass
Vlada [557]
1) 15 / 12 = 1.25 ratio
2) to increase acceleration  1.25 times (with same F, or same engine) you have to lower mass 1.25 times
3) 1515/1.25 = 1212 kg

choose A

6 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
Define moment of momentum. at which condition is it's magnitude zero?​
ololo11 [35]

Let's start with the concept of momentum. What is it? Linear momentum in physics is mathematically written as a product of mass and velocity of an object. Now let us suppose a body of mass m is moving in an inertial frame of reference with velocity v. Consider the fact that no external force is acting on the system. The momentum of this body is given by mv, where m is the mass and v is its velocity. In case of simple real world problems not delving into the realms of relativity, mass is a conserved quantity and it cannot be zero. Hence the velocity of the body must be zero and hence the momentum.

However, photons are considered to have a rest mass zero.

However note the point carefully "rest mass". A body in motion cannot have mass to be zero.

<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em>

7 0
2 years ago
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