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Anuta_ua [19.1K]
3 years ago
14

Jimmy is standing 300 feet away from a rocket that is being shot up into the air away from him at an angle of elevation of 70 de

grees. Jimmy’s line of sight to the rocket has an angle of elevation of 35 degrees. If he is standing 300 feet from the rocket what is the vertical height of the rocket?

Physics
1 answer:
ira [324]3 years ago
4 0

Answer: Vertical height of rocket is 824.24ft

Explanation: see attachment.

Opp/ Adj = Tan 70°

h/ 300 = Tan 70°

h= 300tan 70°

h = 824.24ft

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What is the mass of metal if it has a density of 12.459 hg/cm^3 and displaces 28.7cm^3 of water ​
natima [27]

Answer:

357.6g

Explanation:

Given parameters:

Density  = 12.459g/cm³

Volume of metal  = 28.7cm³

Unknown:

Mass of metal = ?

Solution:

The density of a substance is its mass per unit volume.

To find the mass;

     Mass of metal  = density x volume

Now insert the parameters and solve;

    Mass of metal  = 12.459 x 28.7  = 357.6g

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What is a boyfriend to you? And what age is appropriate to have one?
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7 0
3 years ago
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In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite
Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass_{Hank × Acceleration_{Hank = Mass_{Henry × Acceleration_{Henry

so

Mass_{Hank /  Mass_{Henry = Acceleration_{Henry / Acceleration_{Hank

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass_{Hank /  Mass_{Henry = 1 / 1.26

Mass_{Hank /  Mass_{Henry = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0
3 years ago
What is the kinetic energy of a 50-kg child running to catch the school bus at
Vinvika [58]

Answer:

Option C

100 J

Explanation:

Kinetic energy, KE is given by

KE=0.5mv^{2} where m is the mass and v is the velocity

Substituting 50 Kg for mass, m and 2 m/s for velocity v then we obtain

KE=0.5*50*2^{2}=100 J

Therefore, the child's kinetic energy is equivalent to 100 J

6 0
3 years ago
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