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sasho [114]
3 years ago
6

Model rocket A is launched at an angle of 70° above the x axis, with an initial velocity of 40 m/s. An identical rocket B is lau

nched under the same conditions at an angle of 90° above the x axis, also with an initial velocity of 40 m/s. Which is the true statement about the horizontal distances the rockets will cover?
A. Rocket A will travel farther horizontally than rocket B.
B. Rocket B will travel farther horizontally than rocket A.
C. The rockets will travel the same horizontal distances but will do so in different amounts of time.
D. There is not enough information to answer this question.
Physics
1 answer:
amid [387]3 years ago
3 0
<span>A. Rocket A will travel farther horizontally than rocket B.

This is because from the x axis, 40 m/s at 90 degrees travels directly vertical. 40 m/s at 70 degrees is slightly horizontal, so it will travel further horizontally.</span>
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How is water vapor formed
Bas_tet [7]

Answer:

Y'now water vapor isn't half bad. It's like a vape but it can't affect your lungs. Anyway, Water vapor is water in gaseous instead of liquid form. It can be formed either through a process of evaporation or sublimation.

4 0
3 years ago
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Ubes have sides of equal lengths. What is the fastest way to measure the volume of a cube-shaped table?
Kryger [21]

The quickest technique to calculate the volume of a cube-shaped table is to use a ruler to measure one side, then multiply that figure by three. Option B is correct.

<h3 /><h3>How do you calculate the volume of a cube?</h3>

Assume the side length of the cube under consideration is L units. The volume of the cube is then equal to L³ cubic units.

The volume of a cube is;

V = L³

Ubes have equal-length sides. The quickest technique to calculate the volume of a cube-shaped table is to use a ruler to measure one side, then multiply that figure by three.

Hence option B is correct.

To learn more about the volume of a cube refer

brainly.com/question/26136041

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7 0
2 years ago
the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr
Dmitry [639]

Answer:

The approximate number of decays  this represent  is  N= 23*10^{10}  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is I_a = 2.28 \ mSv

     The source of the radiation is S = 5.49 MeV \ alpha \ particle

 Generally

            1 \  J/kg = 1000 mSv

   Therefore  2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg

Also  1eV = 1.602 *10^{-19}J

  Therefore  2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg}  * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        1 kg ------> 1.423*10^{16}ev \\88.7kg  --------> x

Cross-multiplying and making x the subject

           x = 88.7 * 1.423*10^{16} eV

              x = 126.2*10^{16}eV

Therefore the total  energy  deposited is x = 126.2*10^{16}eV

The approximate number of decays  this represent  is mathematically evaluated as

            N = \frac{x}{S}

Where n is the approximate number of decay

   Substituting values

                             N = \frac{126 .2*10^{16}}{5.49*10^6}  

                                  N= 23*10^{10}  

                     

             

7 0
3 years ago
An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ. Find an expression for
VikaD [51]
We know that
g = LcosΘ 
<span>where g, L and Θ are centripetal gravity length, and angle of object
</span><span>ω² = g/LcosΘ </span>
<span>ω = √(g / LcosΘ) </span>
8 0
3 years ago
Read 2 more answers
A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu
lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

a)

  • While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
  • This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
  • So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

       F_{frk} = -\mu_{k} * F_{n} (1)

       where μk is the kinetic friction coefficient, and Fn is the normal force.

  • Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and  the weight Fg, downward), we conclude that both must be equal and opposite each other:

      F_{n} = F_{g} = m*g (2)

  • We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

       F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)

b)

  • According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
  • In this case, this net force is the friction force which we have already found in a).
  • Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
  • We can write the expression for a as follows:

        a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2  (4)

c)

  • Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

        a = \frac{-v_{o} }{t} (5)

  • Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

       t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)

d)

  • From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
  • So, we can use the following kinematic equation in order to find the displacement before coming to rest:

        v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x  (7)

  • Since the puck comes to a stop, vf =0.
  • Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

       \Delta x  = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m  (8)

e)

  • The total work done by the friction force on the object , can be obtained in several ways.
  • One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
  • Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2}   = -12.4 J  (9)

f)

  • By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
  • P_{Avg} = \frac{\Delta E}{\Delta t}  (10)
  • If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

       P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)

g)

  • The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
  • Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

       P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)

7 0
3 years ago
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