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cluponka [151]
4 years ago
5

A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.

Physics
1 answer:
Anuta_ua [19.1K]4 years ago
5 0

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

         = 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = 0.5mv^{2}

           = 0.5 x 25 x 5.6^{2}  = 392 J

(C) average frictional force = \frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}

  • change in KE (ΔKE) = initial KE - final KE
  • ΔKE = 0.5mv^{2} - 0.5mvf^{2}            
  • when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes  ΔKE = 0.5x25x5.6^{2} - 0 = 392 J

 \frac{-(ΔKE+ΔU)}{h}[/tex] = \frac{-(392 + (-2940))}{12}

=  \frac{(-392 + 2940)}{12} = 212.33 N

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Which is true of planetary orbit A. They travel in elliptical orbits with the Sun in the center B. The speed is the greatest whe
Oksi-84 [34.3K]

Answer:

B is true

Explanation:

This question seems to be incomplete or wrongly written, anyway B is the only one that is true.

Planetary orbits have kinetic and potential energy, the sum of these remains about constant. When closest to the Sun the planet has the lowest potential energy, at this point kinetic energy is maximum, and since speed is related to kinetic energy

Ec = \frac{1}{2} * m * v^2

at maximum kinetic energy speed is maximum too.

A is false because while planets do travel in elliptical orbits, the Sun is not at the center but in one of the focuses.

C is false, you can check the orbital speeds of any of the outer planets. They are slower because at the heigt the orbit there is a smaller gravitational pull. SInce the gravitational pull is the centripetal force, and orbital speed depends on this force, their speed is slower.

8 0
3 years ago
A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocit
Alecsey [184]

Answer:

  F = - 3.53 10⁵ N

Explanation:

This problem must be solved using the relationship between momentum and the amount of movement.

          I = F t = Δp

To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio

        v = d / t

        t = d / v

Reduce SI system

          m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg

          d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m

Let's calculate

         t = 50 10⁻³ / 600

         t = 8.33 10⁻⁵ s

With this value we use the momentum and momentum relationship

        F t = m v - m v₀

As the bullet bounces the speed sign after the crash is negative

       F = m (v-vo) / t

       F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵

       F = - 3.53 10⁵ N

The negative sign indicates that the force is exerted against the bullet

5 0
3 years ago
How do you add two or more than two vectors graphically? please explain in brief..​
Murljashka [212]

Answer:

The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector.

4 0
3 years ago
Read 2 more answers
When rockets split do they come back to earth?
attashe74 [19]
I'm not 100% sure, but I believe what you mean is when they eject the old propulsion motors.  Yes, they land in the ocean and the US Navy retrieves them for later use.
4 0
3 years ago
What is the value of the composite constant (Gme,/r2e) to be multiplied by the mass of the object mo, in equation below:
Sedbober [7]

To solve this problem we will apply the definitions given in Newtonian theory about the Force of gravity, and the Force caused by weight. Both will be defined below, and in equal equilibrium condition to clear the variable concerning acceleration due to gravity. Finally, with the values provided in the statement, it will be replaced.

The equation for the gravitational force between the Earth and the object on the surface of the Earth is

F_g = \frac{Gm_em_o}{r^2_e}

Where,

G = Universal gravitational constant

m_e = Mass of Earth

r_e= Distance between object and center of earth

m_o= Mass of Object

The equation for the gravitational pulling force on the object due to gravitational acceleration is

F_g = m_o g

Equation the two expression we have

m_o g = \frac{Gm_em_o}{r_e^2}

g = \frac{Gm_e}{r_e^2}

This the acceleration due to gravity which is composite constant.

Replacing with our values we have then

g = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5.98*10^{24}kg)}{6378km(\frac{10^3m}{1km})^2}

g = 9.8m/s^2

The value of composite constant is 9.8m/s^2. Here, the composite constant is nothing but the acceleration due to gravity which is constant always.

8 0
4 years ago
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