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weeeeeb [17]
2 years ago
12

A jet plane flying 600 m/s experiences an acceleration of 4.0 g when pulling out of the circular section of a dive. What is the

radius of curvature of this section of the dive?A) 7100 m
B) 650 m
C) 9200 m
D) 1200 m
Physics
1 answer:
Brilliant_brown [7]2 years ago
5 0

To solve this problem we will apply the concepts related to the equations of linear motion and angular motion in order to find the radius. Our values are given as,

g=9.81m/s^2

Then

4g= 39.24 m/s^2

The relation between the acceleration and the angular velocity and the radius is,

a_c=\omega^2*r

The angular velocity and the lineal velocity can be related as,

v=\omega r

The acceleration and the velocity was given, then

39.24=\omega^2 r

600=\omega r \rightarrow r = \frac{600}{\omega}

Replacing at the first equation we have,

39.24 = \omega^2 (\frac{600}{\omega} )

\omega = \frac{39.24}{600}

\omega = 0.0654rad/s

Now using this expression to find the radius we have that

r = \frac{600}{\omega}

r = \frac{600}{0.0654}

r = 9174.31 \approx 9200m

Therefore the correct answer is C.

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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
levacccp [35]

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

3 0
3 years ago
Compare the gravitational force on a 33 kg mass at the surface of the Earth (with ra-
poizon [28]
The formula we will be using is;F = G m1 m2 / r^2 
F earth = 308N 
F moon = G me m2 / (81.3 * 0.27^2 RE^2) = 1/5.927 G me m2 / RE^2 = F earth / 5.927 = 52 N is the force of the earth
So the answer in this question is 52 N.
4 0
3 years ago
1. How much energy would be required to melt 450 grams of ice at 0°C?
xenn [34]

Answer:

Explanation:

1. The amount of heat needed to melt ice at 0°C is equal to the mass of the ice times the latent heat of fusion.

q = mL

q = (450 g) (334 J/g)

q = 150,300 J

q = 150 kJ

2. The amount of heat released by the condensation of steam at 100°C is equal to the mass of the steam times the latent heat of vaporization.

q = mL

q = (325 g) (2260 J/g)

q = 734,500 J

q = 735 kJ

3. q = mL

q = (85 g) (2260 J/g)

q = 192,100 J

q = 190 kJ

4. q = mL

q = (225 g) (334 J/g)

q = 75,150 J

q = 75.2 kJ

5. Above 100°C, water is steam.  The amount of heat needed to increase the temperature of steam is equal to its mass times its specific heat times the change in temperature.

q = mCΔT

q = (20.0 g) (2.03 J/g/°C) (303.0°C − 283.0°C)

q = 812 J

6. q = mCΔT

q = (15.0 g) (2.03 J/g/°C) (250.0°C − 275.0°C)

q = -761 J

7. q = mCΔT

q = (10.0 g) (0.90 J/g/°C) (55°C − 22°C)

q = 297 J

8. q = mCΔT

198 J = (55.0 g) C (15°C)

C = 0.24 J/g/°C

9. q = mCΔT

41,840 J = m (4.184 J/g/°C) (28.5°C − 22.0°C)

m = 1540 g

10. q = mCΔT

q = (193 g) (2.46 J/g/°C) (35°C − 19°C)

q = 7600 J

11. First, the temperature of the ice must be raised to 0°C.

q = mCΔT

q = m (2.09 J/g/°C) (0°C − (-23.0°C))

q/m = 48.1 J/g

Next, the ice must be melted.

q = mL

q/m = 334 J/g

Then, the water must be heated to 100°C.

q = mCΔT

q = m (4.184 J/g/°C) (100°C − 0°C)

q/m = 418.4 J/g

The water is then vaporized.

q = mL

q/m = 2260 J/g

Finally, the steam is heated to its final temperature.

q = mCΔT

q = m (2.03 J/g/°C) (118°C − 100°C)

q/m = 36.5 J/g

So the total amount of energy needed is:

q/m = 48.1 J/g + 334 J/g + 418.4 J/g + 2260 J/g + 36.5 J/g

q/m = 3100 J/g

3 0
3 years ago
White light, with frequencies ranging from 4.00 x 10^14 Hz to 7.90 x 10^14 Hz, is incident on a barium surface. Given that the w
REY [17]

Answer:

0.7515875 eV

4\times 10^{14}\leq f

Explanation:

f = Maximum frequency = 7.9\times 10^{14}\ Hz

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

W = Work function = 2.52 eV

Converting to Joules

W=2.52\times 1.6\times 10^{-19}\\\Rightarrow W=4.032\times 10^{-19}\ J

Maximum photon energy is given by

E=hf\\\Rightarrow E=6.626\times 10^{-34}\times 7.9\times 10^{14}\\\Rightarrow E=5.23454\times 10^{-19}\ J

Maximum Kinetic energy is given by

K=E-W\\\Rightarrow K=5.23454\times 10^{-19}-4.032\times 10^{-19}\\\Rightarrow K=1.20254\times 10^{-19}\ J

Converting to eV

1.20254\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}=0.7515875\ eV

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

W=hf\\\Rightarrow f=\frac{W}{h}\\\Rightarrow f=\frac{4.032\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=6.08512\times 10^{14}\ Hz

The range of frequencies for which no electrons are ejected is

4\times 10^{14}\leq f

5 0
2 years ago
A 0.25 kg steel ball is tied to the end of a string and then whirled in a vertical circle at a constant speed v. The length of t
TiliK225 [7]

Answer:

Explanation:

Let the tension in the string be T . At the top of the circle , total force acting on them = T + mg . This will provide centripetal force

T + mg = m v² / r

4 + .25 x 9.8 = .25 x v² / .62

6.45 = .25 v² / .62

v² = 16

v = 4 m /s .

7 0
2 years ago
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