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torisob [31]
3 years ago
13

In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de

grees? A. 15 degrees. B. 65 degrees. C. 70 degrees. D. 80 degrees. E. 90 degrees.
Physics
1 answer:
snow_tiger [21]3 years ago
5 0

As we know that range of the projectile motion is given by

R = \frac{v^2 sin(2\theta)}{g}

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

\theta, 90 - \theta

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

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Please help ASAP!!
inessss [21]

Answer:

at t=46/22, x=24 699/1210 ≈ 24.56m

Explanation:

The general equation for location is:

x(t) = x₀ + v₀·t + 1/2 a·t²

Where:

x(t) is the location at time t. Let's say this is the height above the base of the cliff.

x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0

v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.

a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².

Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.

Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²

Stone: x(t) = 0 + 22·t - 1/2*9.8 t²

Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:

46 = 22·t

so t = 46/22 ≈ 2.09

Put this t back into either original (i.e., with the quadratic term) equation and get:

x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m

4 0
3 years ago
The mass of earth 6*10^24kg and radius is 400kg Now find the value of acceleration due to gravity when a object is 3600km from t
vlada-n [284]
I don’t worry wewwwww it is a good time to get it done lol lol i don’t worry about it lol lol i lol
5 0
2 years ago
You make an observation that two objects attract. Which object below indicates the claim(s) that could be supported by this obse
wel
4. All of the above.





4 0
3 years ago
A merry-go-round with a a radius of R = 1.9 m and moment of inertia I = 209 kg-m2 is spinning with an initial angular speed of ω
RideAnS [48]

Answer:

340.67 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.9 m

I = Moment of inertia = 209 kgm²

\omega_i = Initial angular velocity = 1.63 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.8 m/s

Initial angular momentum is given by

L=I\omega_i\\\Rightarrow L=209\times 1.63\\\Rightarrow L=340.67\ kgm^2/s

The initial angular momentum of the merry-go-round is 340.67 kgm²/s

8 0
3 years ago
(a) Consider a fusion generator built to create 3.00 GW of power. Determine the rate of fuel burning in grams per hour if the DT
Stels [109]

The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 ×10^1^3 J/g per hour

<h3>How is the rate of fuel burning in grams per hour calculated when the D-T reaction is used?</h3>
  • D + T → He + n
  • The D-T fusion reaction results in a Helium (He) and  neutron (n)

E = 17.59 MeV

Mass = 2.014u + 3.016u

= 5.030u

Energy per Kg = (17.59×10^6×1.6×10^-^1^9) ÷ ( 5.030×1.66×10^-^2^7)

= 3.37×10^1^4 J/Kg

= 3.0× 10^9 J/g

Rate of fuel burning in grams per hour = 3.0× 10^9 ×  3600

= 3.6×3.0×10^1^2

= 1.08 ×10^1^3 J/g per hour

To learn more about fusion reactor and energy production, refer

brainly.com/question/13399644

#SPJ4

4 0
2 years ago
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