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LenKa [72]
3 years ago
6

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant

speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine:
Physics
1 answer:
levacccp [35]3 years ago
3 0

This question is incomplete, the complete question is;

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).

Answer:

the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

Explanation:

Given the data in the question and image below and as illustrated in the second image;

distance S = 40 m

V_B = 54 km/hr

V_A = 72 km/hr

α = 100 m

now, angular velocity of Bxy will be;

ω_B = V_B / α

so, we substitute

ω_B = ( 54 × 1000/3600) / 100

ω_B = 15 / 100

ω_B = 0.15 rad/s

Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

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A type of friction that occurs when air pushes against a moving object causing it to negatively accelerate
Romashka [77]

Answer:

Air resistance

Answer B is correct

Explanation:

The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.

hope this helps

brainliest appreciated

good luck! have a nice day!

6 0
3 years ago
Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy
Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

P_{1}V_{1}=nRT

V_{1}=\dfrac{nRT}{P_{1}}

Put the value into the formula

V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}

V_{1}=2.62\times10^{-3}\ m^3

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}

Put the value into the formula

T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}

T_{2}=233.7\ K

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

P_{1}V_{1}=P_{2}V_{2}

V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}

Put the value into the formula

V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}

V_{2}=0.00786\ m^3

V_{2}=7.86\times10^{-3}\ m^3

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}

V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}

V_{2}=0.0057\ m^3

V_{2}=5.7\times10^{-3}\ m^3

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

6 0
3 years ago
The 480 g bar is rotating as shown what is the angular momentum of the bar about the axle?
Greeley [361]
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:

Angular momentum L = Iω, where 
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>

<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>

<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>

<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>

<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
4 0
3 years ago
What do you mean by pressure​
noname [10]
Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed.
P=F/A
8 0
3 years ago
Read 2 more answers
A car travels along a straight line at a constant speed of 60.0 mi/h for a distance d and then another distance d in the same di
Makovka662 [10]

Answer:

velocity during second d = 20.0 mi/h

Explanation:

Total distance travelled is 2d, with an average velocity of 30.0 mi/h you can express the time travelled in terms of d:

distance = velocity * time

time = distance / velocity

time = 2d/30.0

The time needed for the first d at 60.0 is:

time = d/60.0

The time in the second d you can get it by substracting both times (total time - time for the first d)

second d time = 2d/30.0 - d/60.0

= 4d/60.0 - d/60.0

= 3d/60.0

and with the time (3d/60.0) and the distance travelled (d) you can get the velocity:

velocity = distance / time

velocity = d / (3d/60.0)

= 60.0/3 = 20.0 mi/h

8 0
3 years ago
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