student conducted an experiment and find the density of an ICEBERGE. A students than recorded the following readings. Mass 425 25   g Volume 405 15   mL What experimental value should be quoted for the density of the ICEBERG? Compare your answer with the density of water, which is 3 1.00 10  kg . Show any calculations necessary to justify your answer

Answer:

The initial angular speed is [tex]w_i = 48 \ rad/s[/tex]

Explanation:

From the question we are told that

   The angular displacement is  [tex]\theta = 220 \ rad[/tex]

    The angular speed is  [tex]w_f = 92 \ rad/s[/tex]

    The acceleration is  [tex]\alpha = 14 \ rad/s^2[/tex]

Generally the initial angular speed can be evaluated as

     [tex]w_f ^2 = w_i ^2 + 2 * \alpha * \theta[/tex]

=>  [tex]w_i ^2 = w_f ^2 - 2 * \alpha * \theta[/tex]

substituting values

=>     [tex]w_i ^2 = 92 ^2 - 2 * 14 * 220[/tex]

=>      [tex]w_i ^2 = 2304[/tex]

=>     [tex]w_i = 48 \ rad/s[/tex]

Answer:

FInal speed (v) = 0.509 m/s (Approx)

Explanation:

Given:

Mass of ant (m) = 12 mg

Force (f) = 47 N

Time taken (t) = 0.13 ms

Find:

FInal speed (v) = ?

Computation:

Initial velocity (u) = 0

Impulse = change in momentum

Force × TIme = change in momentum

47 × 0.13 = mv - mu

6.11 = 12 (V)

FInal speed (v) = 0.509 m/s (Approx)

the answer is 1.5 hope this helps

Answer:

1.5

Explanation:

0.5=σ/(15/0.5)

σ=3/2 or 1.5

Answer:

t = 25.5 min

Explanation:

To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.

[tex]t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h[/tex]

Next, you calculate the difference between both times t1 and t2:

[tex]\Delta t=t_1-t_2=2.30h-1.875h=0.425h[/tex]

This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:

[tex]0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s[/tex]

hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph

Complete question is;

A dart is inserted into a spring - loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared to the first?

Answer:

The second dart leaves the gun two times faster than the first one.

Explanation:

If we assume there was no energy loss during the spring - dart energy transfer, we can easily apply the principle of conservation of energy. So;

Potential energy = kinetic energy

Thus;

½kx² = ½mv²

Making velocity "v" the subject, we have;

v = √(kx²/m)

Since the initial distance is "x", thus initial launching velocity is;

v1 = √(kx²/m)

Since next distance is 2x, thus, second launch velocity is;

v2 = √(k(2x)²/m)

Expanding, we have;

v2 = √(4kx²/m)

v2 = 2√(kx²/m)

Comparing this to the one gotten for v1 earlier, we can see that it is double v1.

So, v2 = 2v1

Hence, The second dart leaves the gun two times faster than the first one.

Answer:

15N

Explanation:

F=ma

m=1500g = 1.5kg

a=10m/s2

1.5×10=15 N

Answer:15000gms^-2

Explanation:

F=m×a

m=1500g, a=10ms^-2

F=(1500×10)gms^-2

F=15000gms^-2

Answer:

48 Nm

Explanation:

Moment, or torque, is the cross product of radius and force vectors.

τ = r × F

τ = (0.80 m) (60 N)

τ = 48 Nm

Answer:

a

The free body diagram is shown on the first uploaded image

b

The tension on the rope is  [tex]T=39.16 \ N[/tex]  

Explanation:

From the  question we are told that

    The mass of the box is  m

    The tension on the box is  T

     The angle at which it is pulled is  [tex]\theta = 40^o[/tex]

     The force produced by the strong head wind is [tex]Fw = 30 \ N[/tex]

At equilibrium the net force acting on the block along the horizontal axis is zero i.e

     [tex]Tcos \theta -F_w = 0[/tex]

substituting values

     [tex]Tcos (40) -30 = 0[/tex]

     [tex]Tcos (40) = 30[/tex]

     [tex]T(0.76604)) = 30[/tex]

     [tex]T=39.16 \ N[/tex]      

Answer:

1.95 x 10^8 m/s.

Explanation:

Answer:

the answer is 1.95 x 10^8 m/s

Explanation:

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The uncertainty in inverse frequency is  [tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]

Explanation:

From the question we are told that

   The value of the proportionality constant is  [tex]k = 5 \frac{Hz }{T}[/tex]

   The strength of the magnetic field is  [tex]B = 20 \ T[/tex]

   The change in this strength of magnetic field is  [tex]\Delta B = 3 \ T[/tex]

The magnetic field is given as

           [tex]B = \frac{k}{\frac{1}{w} }[/tex]

Where [tex]w[/tex] is frequency

The uncertainty or error of the field is given as

         [tex]\Delta B = \frac{k }{[\frac{1}{w}^]^2 } \Delta [\frac{1}{w} ][/tex]

The uncertainty in inverse frequency is given  as

           [tex]\Delta [\frac{1}{w} ] = \frac{\Delta B}{k [\frac{1}{w^2} ]}[/tex]

                    [tex]\Delta [\frac{1}{w} ]= \frac{\Delta B}{k (B)^2 }[/tex]

substituting values

                  [tex]\Delta [\frac{1}{w} ]= \frac{3}{5 (20)^2 }[/tex]

               [tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]

Answer:

a

The total initial momentum of the two-block system is  [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]

b

The magnitude of the final velocity of the two-block system [tex]v_f = 1.9014 \ m/s[/tex]

c

 the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is  

    [tex]\Delta KE =- 847.08 \ J[/tex]

Explanation:

From the question we are told that

    The mass of first  block  is [tex]m_1 = 2.50 \ kg[/tex]

      The initial velocity of first   block is [tex]u_1 = 27.0 \ m/s[/tex]

          The mass of second block is  [tex]m_2 = 33.0\ kg[/tex]

          initial velocity of second block is  [tex]u_2 = 0 \ m/s[/tex]

The magnitude of the of the total initial momentum of the two-block system is mathematically repented as

        [tex]p_i = (m_1 * u_1 ) + (m_2 * u_2)[/tex]

substituting values

        [tex]p_i = (2.50* 27 ) + (33 * 0)[/tex]

        [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]

According to the law of linear momentum conservation

        [tex]p_i = p_f[/tex]

Where  [tex]p_f[/tex] is the total final momentum of the system which is mathematically represented as

       [tex]p_f = (m_+m_2) * v_f[/tex]

Where [tex]v_f[/tex] is the final velocity of the system

      [tex]p_i = (m_1 +m_2 ) v_f[/tex]

substituting values

       [tex]67.5 = (2.50+33 ) v_f[/tex]

        [tex]v_f = 1.9014 \ m/s[/tex]

The change in kinetic energy is mathematically represented as

     [tex]\Delta KE = KE_f -KE_i[/tex]

Where [tex]KE_f[/tex] is the final kinetic energy of the two-body system  which is mathematically represented as

        [tex]KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2[/tex]

substituting values

        [tex]KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2[/tex]

        [tex]KE_f =64.17 J[/tex]

While [tex]KE_i[/tex] is the initial kinetic energy of the two-body system

     [tex]KE_i = \frac{1}{2} * m_1 * u_1^2[/tex]

substituting values

       [tex]KE_i = \frac{1}{2} * 2.5 * 27^2[/tex]

        [tex]KE_i = 911.25 \ J[/tex]

So

    [tex]\Delta KE = 64.17 -911.25[/tex]

  [tex]\Delta KE =- 847.08 \ J[/tex]

Answer:

A. 52 min

.A. 47 watts

Explanation:

Given that;

jim weighs 75 kg

and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.

Using the following relation to determine the amount of calories burned per minute while walking; we have:

[tex]\dfrac{MET*weight (kg)*3.5}{200}[/tex]

here;

MET = energy cost of a physical activity for a period of time

Obtaining the data for walking with a speed of 3.3 mph From the  standard chart for MET, At 3.3 mph; we have our desired value to be 4.3

However;

the calories burned in a minute = [tex]\dfrac{4.3*75 (kg)*3.5}{200}[/tex]

= 5.644

Therefore, for walking for 52 mins; Jim  burns approximately 293.475 kcal which is nearest to 300 kcal.

4.

Given that:

mass m = 75 kg

intensity = 6 kcal/min

The eg ergometer work rate = ??

Applying the formula:

[tex]V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7[/tex]

where ;

[tex]V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}[/tex]

[tex]V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}[/tex]

[tex]V_O_2 ( intensity ) = 0.0012[/tex]

∴[tex]0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min[/tex]

Converting to watts;

Since;  6.118kg-m/min is =  1 watt

Then 291.66 kgm /min will be equal to 47.67 watts

≅ 47 watts

Answer:

a) The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex], b) The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].

Explanation:

This question is incomplete and complete version will be presented herein:

A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring (b) Find the angular frequency of the block 's vibrations.

a) Since spring is hung from the ceiling and is stretched by action of gravity on 0.573 kilogram block. According to the Hooke's Law, force experimented by the spring is directly proportional to elongation. An expression describing the phenomenon is presented and described below: (System at equilibrium - Newton's Second Law)

[tex]m\cdot g = k\cdot \Delta x[/tex]

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravitational constant, measured in meters per square second.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]\Delta x[/tex] - Spring linear deformation, measured in meters.

Now, the spring constant is cleared in this equation and outcome is computed: ([tex]m = 0.573\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta x = 0.198\,m[/tex])

[tex]k = \frac{m\cdot g}{\Delta x}[/tex]

[tex]k = \frac{(0.573\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.198\,m}[/tex]

[tex]k = 28.381\,\frac{N}{m}[/tex]

The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex].

b) Let suppose that mass-spring system is experimenting a simple harmonic motion, so that angular frequency is equal to:

[tex]\omega = \sqrt{\frac{k}{m} }[/tex]

Given that [tex]k = 28.381\,\frac{N}{m}[/tex] and [tex]m = 0.573\,kg[/tex], the angular frequency, measured in radians per second, of the block is:

[tex]\omega = \sqrt{\frac{28.381\,\frac{N}{m} }{0.573\,kg} }[/tex]

[tex]\omega = 7.038\,\frac{rad}{s}[/tex]

The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].

Answer:

2.95m

Explanation:

The farthest distance the object can move is the radius of the circle of which the Simple harmonic motion is assumed to be a part

But V = w× r; where V is velocity,

w is angular velocity and r is radius.

Also,

a= w2r; where a is linear acceleration

but a = v× r ; by comparing both equations

Hence r = a/v =8.6/2.45 =3.51m

But the horizontal distance of the motion is given by:

X = rcosx ; where x is the angle

X is the distance covered.

We know that the maximum value of cos x is 1 which is 0°

When the object moves in a fashion directly parallel to an horizontal distance, maximum distance would be reached and hence:

X = r=3.51m

Meaning the object needs to travel 3.51-0.56=2.95m further.

Note: the acceleration of the motion is constant whether it is swinging towards the left or right.

When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left and the amplitude of motion A = 0.732 m.

The distance between the central and extreme points for a moving particle is known as the amplitude of motion.

The given data to find the amplitude of motion,

Object displaced = 0.560 m

Velocity = 2.45 m/s

Acceleration = 8.60 m/s²

Starting with sine:

x(t) = Asin(ωt)

so that t = 0, x = 0

x(t) = 0.56 m = Asin(ωt)

v(t) = x(t)'= 2.45 m/s = Aωcos(ωt)

a(t) = v(t)'= -8.60 m/s² = -Aω²sin(ωt)

x(t) / a(t) = Asin(ωt) / -Aω²sin(ωt)

0.56m / -8.60 m/s² = -1 / ω²

ω² = 15.3571 rad^2/s^2

ω = 3.91881 rad/s  

x(t) / v(t) = Asin(ωt) / Aωcos(ωt)

0.560m / 2.45m/s = tan(3.91t) / 3.91rad/s

0.8937= tan(3.91t)

t = 0.176 s  

x(0.176) = Asin(3.59×0.176)

0.65 m= Asin(0.631)

A = 0.732 m is the amplitude of motion.

To know more about Amplitude of motion,

https://brainly.com/question/12967589

#SPJ2

Answer:

The angular momentum is  [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]

Explanation:

From the question we are told that

    The mass of the woman is  [tex]m = 50 \ kg[/tex]

     The angular  speed of the rim is  [tex]w = 0.80 \ rev/s = 0.8 * [\frac{2 \pi}{1} ] = 5.024 \ rad \cdot s^{-1}[/tex]

      The mass of the disk is  [tex]m_d = 110 \ kg[/tex]

       The radius of the disk is [tex]r_d = 4.0 \ m[/tex]

The moment of inertia of the disk is mathematically represented as

        [tex]I_D = \frac{1}{2} m_d r^2_d[/tex]

substituting values

          [tex]I_D = \frac{1}{2} * 110 * 4^2[/tex]

          [tex]I_D = 880 \ kg \cdot m^2[/tex]

The moment of inertia of the woman is  

          [tex]I_w = m * r_d^2[/tex]

substituting values

        [tex]I_w = 50 * 4^2[/tex]

       [tex]I_w =800\ kg[/tex]

The moment of inertia of the system (the woman + the large disk ) is  

        [tex]I_t = I_w + I_D[/tex]

substituting values  

      [tex]I_t = 880 +800[/tex]

     [tex]I_t =1680 \ kg \cdot m^2[/tex]

The angular momentum of the system is

      [tex]L = I_t w[/tex]

substituting values  

      [tex]L = 1680 * 5.024[/tex]

      [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]

Energy = (power) x (time)

-- For the toaster:

Power = 1.4 kW  =  1,400 watts

Time = 5.4 minutes = 324 seconds

Energy = (1,400 W) x (324 s)  =  453,600 Joules

-- For the CFL bulb:

Power = 11 watts

Time = 10.5 hours = 37,800 seconds

Energy = (11 W) x (37,800 s)  =  415,800 Joules

-- The toaster uses energy at 127 times the rate of the CFL bulb.

-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.

-- The toaster is used for 0.0086 times as long as the CFL bulb.

-- The CFL bulb is used for 116.7 times as long as the toaster.    

-- The toaster uses 9.1% more energy than the CFL bulb.

-- The CFL bulb uses 8.3% less energy than the toaster.  

Answer:

7.2 m/s

Explanation:

The maximum speed is the amplitude times the frequency.

v = Aω

v = (0.76 m / 2) (180 rev × 2π rad/rev / 60 s)

v = 7.2 m/s

Answer:

v = 7.67 m/s

Explanation:

The equation for apparent weight in the situation of weightlessness is given as:

Apparent Weight = m(g - a)

where,

Apparent Weight = 360 N

m = mass passenger = 61.2 kg

a = acceleration of roller coaster

g = acceleration due to gravity = 9.8 m/s²

Therefore,

360 N = (61.2 kg)(9.8 m/s² - a)

9.8 m/s² - a = 360 N/61.2 kg

a = 9.8 m/s² - 5.88 m/s²

a = 3.92 m/s²

Since, this acceleration is due to the change in direction of velocity on a circular path. Therefore, it can b represented by centripetal acceleration and its formula is given as:

a = v²/r

where,

a = centripetal acceleration = 3.92 m/s²

v = speed of roller coaster = ?

r = radius of circular rise = 15 m

Therefore,

3.92 m/s² = v²/15 m

v² = (3.92 m.s²)(15 m)

v = √(58.8 m²/s²)

v = 7.67 m/s

Answer:

e. 400 Hz

Explanation:

In closed organ pipe,  only odd harmonics of fundamental note is possible .

The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .

200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc

600 Hz , 1000 Hz , 1400 Hz  , 1800 Hz .

Frequency not possible is 400 Hz .

Answer :the resultant of two vectors can be found using either the parallelogram method or the triangle method. don't know if this was helpful ?

Explanation:

Answer:

Analytical method.

Answer:

W = 414 J, correct is B

Explanation:

Work is defined by

        W = ∫ F .dx

where F is the force, x is the displacement and the point represents the dot product

this expression can also be written with the explicit scalar product

        W = ∫ F dx cos θ

where is the angle between force and displacement

for this case as the force is constant

         W = F x cos θ

calculate

         W = 25.0 18.0 cos (-23)

         W = 414 J

the correct answer is B

Answer:

  1.234567 kA

Explanation:

The prefix k stands for kilo-, or 10³. The prefix m stands for milli-, or 10⁻³. The sum shown is ...

  1.234 kA + 0.000567 kA = 1.234567 kA

Answer:

Air resistance

Answer B is correct

Explanation:

The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.

hope this helps

brainliest appreciated

good luck! have a nice day!

Answer:

Explanation:

Volume of asteroid = 4/3 x π x 160³

= 17.15 x 10⁶

mass = volume x density

= 17.15 x 10⁶ x 2600

= 445.9 x 10⁸ kg

kinetic energy

= 1/2 x 445.9 x 10⁸  x( 12.6 )² x 10⁶

= 35.4 x 10¹⁷ J .

2 )

energy of 15 megaton

= 4.184 x 10¹⁵ x 15 J

= 62.76 x 10¹⁵ J

No of bombs required

= 35.4 x 10¹⁷ / 62.76 x 10¹⁵

= 56.4 Bombs .

Answer:

Explanation:

vertical component of the velocity of arrow

= 26 sin 60 = 22.516 m

height reached by it after 3.99 s

h = ut - 1/2 g t²

= 22.516 x 3.99 - .5 x 9.8 x 3.99²

= 89.83 - 78

11.83 m

Total height of cliff = 1.55 + 11.83

= 13.38 m

c ) maximum height covered s

v² = u² - 2gs

0 = u² - 2gs

s = u² / 2g

= 22.516² / 2 x 9.8

= 25.86

maximum height reached

= 25.86 + 1.55

= 27.41 m

d )

vertical speed after 3.99 s

v = u - gt

= 22.516 - 9.8 x 3.99

= -16.586

Horizontal component will remain unchanged

Horizontal component = 26 cos 60

= 13 m /s

Resultant of two velocities

= √ 13²+ 16.568²

= 21 m /s

Answer:

K = 3.9 kJ

Explanation:

The kinetic energy ([tex]K_{T}[/tex]) added is given by the difference between the final kinetic energy and the initial kinetic energy:

[tex] K_{T} = K_{f} - K_{i} [/tex]  

The initial kinetic energy is:

[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} [/tex]

Where m₁ is the mass of the object before the explosion and v₁ is its velocity

[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} = \frac{1}{2}70 kg*(21 m/s)^{2} = 1.54 \cdot 10^{4} J [/tex]

Now, the final kinetic energy is:

[tex] K_{f} = \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2} [/tex]

Where m₂ and m₃ are the masses of the 2 pieces produced by the explosion and v₁ and v₂ are the speeds of these pieces

Since m₂ is 4 times as massive as m₃ and v₃ = 0, we have:

[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}v_{2}^{2} + \frac{1}{2}*\frac{1}{5}m_{1}*0 [/tex]   (1)          

By conservation of momentum we have:

[tex] p_{i} = p_{f} [/tex]

[tex] m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3} [/tex]  

[tex] m_{1}v_{1} = \frac{4}{5}m_{1}v_{2} + \frac{1}{5}m_{1}*0 [/tex]

[tex] v_{2} = \frac{5}{4}v_{1} [/tex]     (2)

By entering (2) into (1) we have:

[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}(\frac{5}{4}v_{1})^{2} = \frac{1}{2}*\frac{4}{5}70 kg(\frac{5}{4}*21 m/s)^{2} = 1.93 \cdot 10^{4} J [/tex]  

Hence, the kinetic energy added is:

[tex] K_{T} = K_{f} - K_{i} = 1.93 \cdot 10^{4} J - 1.54 \cdot 10^{4} J = 3.9 \cdot 10^{3} J [/tex]  

Therefore, the kinetic energy added to the system during the explosion is 3.9 kJ.

I hope it helps you!

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

[tex]U=(M+m)gh[/tex]       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

[tex]h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m[/tex]

The potential energy is:

[tex]U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J[/tex]

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

[tex]U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}[/tex]

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

[tex]Mv_1+mv_2=(M+m)v[/tex]    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

[tex]M(0)+mv_2=(M+m)v[/tex]    

[tex]v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}[/tex]

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

[tex]I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}[/tex]

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

[tex]T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N[/tex]    

The force on the cord after the impact is 2.59N

Answer:

Explanation:

Apply the law of conversion of energy

[tex]V_f = \sqrt{2gh}[/tex]

where,

h = height of which the bullet and block rise after impact

[tex]h = L - Lcos\theta\\\\h = 2.2 - (2.2*cos60)\\\\h = 1.1m[/tex]

Therefore,

[tex]V_f = \sqrt{2gh}\\\\V_f = \sqrt{2*9.8*1.1}\\\\V_f = 4.64m/s[/tex]

From conservation of momentum principle, [tex]m_Bv_B = 0[/tex]

[tex]m_ov_o + m_Bv_B = (m_b+m_B)V_f\\\\0.006V_o = (0.006+1)*4.64\\\\V_o = 777.97m/s[/tex]

C) The force in the cable is due to the centrfugal force of the system, which is due to the motion of the system is a curved path and weight of the system

[tex]F = \frac{m_b+m_B}{L}V_f^2 + (m_b+m_B)g\\\\F = \frac{0.006+1}{2.2}*4.64^2 + (0.006+1)9.81\\\\F = 19.71N[/tex]

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Answer:

The speed of the freight car decreases.

Explanation:

According to the law of conservation of momentum indicates that for colliding in an isolated system, the total momentum pre and post collision is same for the two objects this is done because the momentum that one item has lost is same for the momentum that the other received

In the given situation, the freight car travels at constant speed along a frictionless railroad line. The top floor freight car is open. Then a huge load of coal is dumped inside the car.

Therefore the speed of the freight car decreased by applying the law of conservation of momentum i

Answer:

1.been both -ve charged or both +be charged particles

2. 3.52mC

Explanation:

For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.

Now the Force sustain by the extended string is

F = Ke;

Where K is the force constant of the string, 320 N/m

e is the extension,0.033 m

F = 320 × 0.033 =10.56N

2.But according to columns law of charge;

F = kQ1 Q2

But Q1=Q2{ since the charge are of the same magnitude}.

Hence F = KQ^2

Where K is columns constant =9×10^9F/m

Hence Q=√F/K

Q= √10.56/9×10^9

=3.52×10^-3C

= 3.52mC