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storchak [24]
3 years ago
9

Please help ASAP! I will give the first answer brainliest!

Physics
1 answer:
Natalija [7]3 years ago
3 0

Answer:

Sorry for being late. Hope this helps

Explanation:

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To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Kryger [21]

Answer:

H = 10.05 m

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

so this is the speed at the top of flag pole

now we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

now the height of flag pole is given as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
3 years ago
What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the cal
ivanzaharov [21]
<span>Answer: q Cu = q H2O so, (121g)(0.385 J/g-K)(100.4C-x) = (150g)(4.18 J/g-K)(x-25.1C) x= 30.3C</span>
5 0
4 years ago
Ask a testable question about each event listed in part a
Ratling [72]

Answer:

What?-

Explanation:

Where’s the story or whatever like that?

3 0
3 years ago
In trial 1 of an experiment, a cart moves with a speed of vo on a frictionless, horizontal track and collides with another cart
marta [7]

Answer:

1) elastic shock, the velocity of the center of mass does not change

2) inelastic shock, he velocity of the mass center   change

Explanation:

The position of the center of mass of your system is defined by

          x_{cm} = \frac{1}{M} \sum x_i m_i

in this case we have two bodies

          x_{cm} = \frac{1}{M} (x₁m₁ + x₂ m₂)

the velocity of the center of mass is

          x_{cm} = dx_{cm} / dt = \frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )

          x_{cm} = \frac{1}{M} ( m_1 v_1 + m_2 v_2 )

where M is the total mass of the system.

Therefore to answer this question we have to find the velocity of the body after the collision.

Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.

Let's solve each case separately.

2) inelastic shock

initial instant. Before the crash

         p₀ = m₁ v₀ + 0

final instant. After the collision with the cars together

        p_f = (m₁ + m₂) v

         p₀ = p_f

         m₁ v₀ = (m₁ + m₂) v

         v = \frac{m_1}{m_1+m_2}  v₀

let's find the velocity of the center of mass

         M = m₁ + m₂

initial.

         v_{cm o} = \frac{1}{m_1 +m_2} (m₁ vo)

final

         v_{cm f} = \frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o ) ( v) = v

         v_{cm f} =  \frac{m_1}{M^2} v_o

Let's find the ratio of the velocities of the center of mass

          vcmf / vcmo = \frac{1}{M} = \frac{1}{m_1 +m_2}

           

           

therefore the velocity of the mass center   change

1) elastic shock

initial instant.

           p₀ = m₁ v₀

final moment

           p_f = m₁ v_{1f} + m₂ v_{2f}

           p₀ = p_f

           m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}

           m₁ (v₀ - v_{2f}) = m₂ v_{2f}

in this case the kinetic energy is conserved

           K₀ = K_f

          ½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

           m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²

           m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}

we write our system of equations

           m₁ (v₀ - v_{1f}) = m₂ v_{2f}             (1)

           m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²

we solve the system

             v₀ + v_{1f} = v_{2f}

we substitute and look for the final speeds

             v_{1f} = \frac{m_1 -m_2}{m1 +m2 } v_o

             v_{2f} = \frac{2 m_1}{m-1+m_2} vo

now let's find the velocity of the center of mass

initial

          v_{cm o} = \frac{1}{M} m₁ v₀

final

          v_{cm f} = \frac{1}{M}  (m₁ v_{1f} + m₂ v_{2f} )

          v_{cm f} = \frac{1}{M} [  m_1  \frac{m_2}{M} + m_2  \frac{2 m_1}{M} ] v₀

          v_{cm f} = \frac{1}{M^2} ( m₁² - m₁m₂ +2 m₁m₂) v₂

          v_{cm f} = \frac{1}{M^2} (m₁² + m₁ m₂) v₀

let's look for the relationship

         v_{cm f} / v_{cm o} = \frac{1}{M} M

         v_{cm f} / v_{cm o} = 1

therefore the velocity of the center of mass does not change

we see in either case the velocity of the center of mass does not change.

4 0
3 years ago
A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And
Naily [24]

Answer:

700J

Explanation:

8 0
3 years ago
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