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Lelu [443]
3 years ago
12

An object has a mass of 24.5 Kilograms and an acceleration of 0 meters per second squared. What would be its first Force in Newt

ons?
A 100 Newtons
B 10.4 Newtons
C 24.5 Newtons
D 0 Newtons
Other ________
Physics
1 answer:
swat323 years ago
4 0
I would say D because I think it’s supposed to be mass times acceleration
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Greg throws a 2.8-kg pumpkin horizontally off the top of the school roof in order to hit Mr. H's car. The car has parked a dista
Igoryamba

Answer:

The horizontal velocity is v = 9.2 m/s

Explanation:

From the question we are told that

     The mass of the pumpkin is  m = 2.8 \ kg

      The distance of the the car from the building's base is  d = 13.4 \ m

       The height of the roof is h = 10.4 \ m

       

The height is mathematically represented as

         h = \frac{1}{2} gt^2

Where g is the acceleration due to gravity which has a value of g =9.8 \ m/s^2

substituting values

          10.4= 0.5 * 9.8 * t

making the time taken the subject of the formula

         t = \frac{10.4}{0.5 * 9.8 }

          t = 1.457 \ s

The speed at which the pumpkin move horizontally can be represented mathematically  as

                         v = \frac{d}{t}

substituting values

                     v =\frac{13.4}{1.457}

                     v = 9.2 m/s

7 0
3 years ago
An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi
Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

                  V_x'=0.8c\\V=0.6c\\m=4*10^5kg

  • Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

                  KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}

  • So, to V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
  • We have an expression from Lorents transformation for relativistic law of addition of velocities as,

                      V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V

  • Substituting values, we get,

          V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c

  • Thus, the KE will be,

              KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

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Choice-C is a correct statement.
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