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3 years ago

14

a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi

ne the spring constant of the gun

1 answer:

Vikki [24]3 years ago

7 0

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

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3 years ago

With what velocity Must a 0.53 kg softball be moving to be equal to the momentum of a 0.31 kg baseball moving at 21 m/s

vladimir2022 [97]

Answer:12.28m/s

Explanation:

momentum of baseball =mass of baseball x velocity of baseball

Momentum of baseball =0.31x21

Momentum of baseball =6.51kgm/s

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Velocity of softball =6.51/0.53

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3 0

4 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 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

In the diagram, q1= +8.0 C, q2= +3.5 C, and q3 = -2.5 C. q1 to q2 is 0.10 m, q2 to q3 is 0.15 m. What is the net force on q2? La

yulyashka [42]

Answer:

f(t) = 28,7 [N]

Explanation: IMPORTANT NOTE: IN PROBLEM STATEMENT CHARGES ARE IN C (COULOMBS) AND IN THE DIAGRAM IN μC. WE ASSUME CHARGES ARE IN μC.

The net force on +q₂ is the sum of the force of +q₁ on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)

The two forces have the same direction to the right of charge q₂, we have to add them

Then

f(t) = f₁₂ + f₃₂

f₁₂ = K * ( q₁*q₂ ) / (0,1)²

q₁ = + 8 μC then q₁ = 8*10⁻⁶ C

q₂ = + 3,5 μC then q₂ = 3,5 *10⁻⁶ C

K = 9*10⁹ [ N*m² /C²]

f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻² [ N*m² /C²]* C*C/m²

f₁₂ = 252*10⁻¹ [N]

f₁₂ = 25,2 [N]

f₃₂ = 9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²

f₃₂ = 78,75*10⁻³/ 2,25*10⁻²

f₃₂ = 35 *10⁻¹

f₃₂ = 3,5 [N]

f(t) = 28,7 [N]

5 0

3 years ago

Read 2 more answers