Calculate the potential energy of 5.0g (0.005 kg) paper airplane 5.0 meters above the ground

. 30

schepotkina [342]

Answer:

Explanation:

Length if the bar is 1m=100cm

The tip of the bar serves as fulcrum

A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum

The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.

Using moment

Sum of the moments of all forces about any point in the plane must be zero.

Let take moment about the fulcrum

100×20-F×2=0

2000-2F=0

2F=2000

Then, F=1000N

The force acting in the crate lid is 1000N

Option D is correct

7 0

2 years ago

A stunt man jumps from the top of a building and lands 10 meters below his initial height. In case A, the stunt man lands on a s

krek1111 [17]

Answer:

Explanation:

The stunt will likely sustain serious injury in case of concrete blocks because the average force acting on the person will be more because concrete blocks do not squeeze to provide more time for the force to act on the body instead it acts for a small amount of interval.

$Impulse=F_{avg}\times \Delta T$

As impulse is constant so time requires to act force on the body is more as compared to concrete block and thus average force in mattress case is less.

8 0

3 years ago

Can someone explain to me #4.

nalin [4]

A) We balance the masses: 4(1.00728) vs 4.0015 + 2(0.00055)4.02912 vs. 4.0026This shows a "reduced mass" of 4.02912 - 4.0026 = 0.02652 amu. This is also equivalent to 0.02652/6.02E23 = 4.41E-26 g = 4.41E-29 kg.
b) Using E = mc^2, where c is the speed of light, multiplying 4.41E-29 kg by (3E8 m/s)^2 gives 3.96E-12 J of energy.
c) Since in the original equation, there is only 1 helium atom, we multiply the energy result in b) by 9.21E19 to get 3.65E8 J of energy, or 365 MJ of energy.

4 0

2 years ago

When a carousel is in motion, the movement of the carousel horse can best be described as?

tatiyna

Supposing the carousel is rotating with constant speed, the movement is uniform angular motion.

5 0

3 years ago

Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the c

Amanda [17]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is $v_A= 4m/s$

Explanation:

From the question we are told that

The length of the string is $L = 1m$

The initial speed of block A is $u_A$

The final speed of block A is $v_A = \frac{1}{2}u_A$

The initial speed of block B is $u_B = 0$

The mass of block A is $m_A = 7kg$ gh

The mass of block B is $m_B = 2 kg$

According to the principle of conservation of momentum

$m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}$

Since block B at initial is at rest

$m_A u_A = m_Bv_B + m_A \frac{u_A}{2}$

$m_A u_A - m_A \frac{u_A}{2} = m_Bv_B$

$m_A \frac{u_A}{2} = m_Bv_B$

making $v_B$ the subject of the formula

$v_B =m_A \frac{u_A}{2 m_B}$

Substituting values

$v_B =\frac{7 u_A}{4}$

This $v__B$ is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of $v__B'$ at the top of the vertical circle

The angular centripetal acceleration would be mathematically represented

$a= \frac{v^2_{B}'}{L}$

Note that this acceleration would be toward the center of the circle

Now the forces acting at the top of the circle can be represented mathematically as

$T + mg = m \frac{v^2_{B}'}{L}$

Where T is the tension on the string

According to the law of energy conservation

The energy at bottom of the vertical circle = The energy at the top of

the vertical circle

This can be mathematically represented as

$\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L$

From above

$(T + mg) L = m v^2_{B}'$

Substitute this into above equation

$\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = T + 5mgL$

The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero

This is mathematically represented as

$\frac{49 mv_A^2}{16} = 5mgL$

making $v_A$ the subject

$v_A = \sqrt{\frac{80mgL}{49m} }$

substituting values

$v_A = \sqrt{\frac{80* 9.8 *1}{49} }$

$v_A= 4m/s$

6 0

3 years ago