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MrRissso [65]
2 years ago
14

A projectile is launched horizontally east at a speed of 29.4 M/s towards a wall 88.2 m away. What is the velocity of the projec

tile at the moment it strikes the wall?
A. 29.4 m/s down
B. 29.4 m/s east
C. 29.4 m/s at 45(degrees) down from horizontal toward the east
D. 41.6 m/s at 45(degrees) down from horizontal toward the east
Physics
1 answer:
Drupady [299]2 years ago
6 0

Time before projectile hits wall

= 88.2 m / 29.4 m/s = 3 seconds

Vertical velocity of projectile after three seconds

= 3*9.8 = 29.4 m/s

Horizontal velocity of projectile after three seconds, assuming no air resistance

= 29.4 m/s  (given)

Conclusion:

velocity of projectile when it hits the wall

= < 29.4, -29.4> m/s

= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal

= 41.58 m/s east-bound at 45 degrees below horizontal.

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valentina_108 [34]

Answer:

3. none of these

Explanation:

The rotational kinetic energy of an object is given by:

K=\frac{1}{2}I \omega^2

where

I is the moment of inertia

\omega is the angular speed

In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

K_1 = \frac{1}{2}I (\omega)^2 = \frac{1}{2}I\omega^2

For disk 2:

K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2

so the total energy is

K=K_1 + K_2 = \frac{1}{2}I\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2

So, none of the options is correct.

5 0
2 years ago
If a helicopter's mass is 4,500kg and the net force on it is 18,000 N upward, what is its acceleration?
Vinvika [58]
The acceleration is0.25m/s^2
4 0
1 year ago
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In a hyrdo electic power plant the moving water turns the turbine creating ____________ energy.
Marrrta [24]

Answer:

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Explanation:

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3 0
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Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person
Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

C=2\pi r    (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

L=Lo[1+\alpha \Delta T]         (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0
1 year ago
A gymnast of mass 70.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
DiKsa [7]

Answer:

T = 686.7N

Explanation:

For this exercise we will use Newton's second law in this case there is no acceleration,

      ∑ F = ma

      T -W = 0

The gymnast's weight is

     W = mg

We clear and calculate the tension

     T = mg

     T = 70 9.81

     T = 686.7N

3 0
1 year ago
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