When a balloon is deflating, why does air leave the balloon?

All points in a rigid body move with the same velocity and acceleration if the rigid body is subjected to pure translational motion.

When a rigid body is subjected to pure translational motion, all points in the body move with the same velocity and acceleration. This means each point in the body experiences the same linear velocity and acceleration at any given instant.

Answer:

The mass of the puck is 0.166 kg

Explanation:

It is given that,

Speed of the ice hockey puck, u = 12 m/s

Impulse delivered by the hockey stick, J = 4 kg-/s

After delivering an impulse, the puck move off in the opposite direction with the same speed, v = -12 m/s

We need to find the mass of the puck. The impulse is equal to the change in momentum i.e.

[tex]J=m(v-u)[/tex]

m is the mass of the puck

[tex]4=m(-12-12)[/tex]

[tex]4=-24\ m[/tex]

[tex]m=\dfrac{1}{6}=0.166\ kg[/tex]

So, the mass of the puck is 0.166 kg. Hence, this is the required solution.

When the mass of the puck that moves with the same speed in opposite direction is equal to 0.1667 kg.

Impulse is a large force that is applied to an object for a very short period of time. It is given by the formula,

[tex]J =m(v-u) = F(\delta t)[/tex]

where v and u are the final and initial velocity, F is the force that is been applied, m is the mass of the object, t is the time for which the force is been applied.

We know that the impulse can be written as,

[tex]J = m(v-u)[/tex]

Given to us

J  =  4 kg⋅m/s,

u = 12 m/s

v = -12 m/s(opposite in direction with the same speed)

Substitute the values,

[tex]4 = m(-12-12)\\\\4 = m(-24) \\\\m = -0.1666\\\\m = 0.1667 \rm\ kg[/tex]

Since the mass of the punk can not be negative, therefore, the mass of the puck is 0.1667 kg.

Hence, when the mass of the puck that moves with the same speed in opposite direction is equal to 0.1667 kg.

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Answer:

A-2

B-1

Explanation:

Cylindrical projection is a type of map projection that projects the earth's surface on a cylindrical tangent. It is useful in projecting the equatorial region and the near-polar region but it gets distorted at the poles. This distortion occurs in terms of both the scale and the shape.

Planar projection is a type of map projection in which the three-dimensional earth's surface is projected on a flat surface. It is more efficient in mapping the polar regions. The North pole or the South Pole in this type of map are located at the center of the map.

Thus, the correct answers are being matched.

To find the energy in joules equivalent to one jelly doughnut, multiply the Calorie value by 4.184. The number of stairs required to perform the same work as the energy in one jelly doughnut is calculated by dividing the work done to move up the stairs by the work done to move one stair height. To account for the body's efficiency in converting energy, divide the number of stairs calculated in part (b) by the efficiency.

(a) To find the number of joules of energy equivalent to one jelly doughnut, we need to convert the Calorie value to joules. One Calorie (capital C) is equal to 1000 calories (lowercase c), which is equal to 4.184 joules. Therefore, one jelly doughnut contains 625 calories × 4.184 joules/calorie = 2615 joules.

(b) The work done by climbing stairs is equal to the product of the force applied and the distance moved. The force can be calculated using the weight of the woman, which is equal to her mass multiplied by the acceleration due to gravity (66 kg × 9.8 m/s2). Dividing the work done to move up the stairs by the work done to move one stair height (15 cm) gives the number of stairs that the woman must climb. This is equal to 1764 joules ÷ (66 kg × 9.8 m/s2 × 0.15 m) = 181 stairs.

(c) Since the human body is only 26% efficient in converting chemical energy to mechanical energy, we need to divide the work calculated in part (b) by the efficiency to account for this loss. Therefore, the number of stairs the woman must climb to work off her breakfast is 181 stairs ÷ 0.26 = 696 stairs.

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Answer:

s = 23.72 m

v = 21.56 m/s²  

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

   u = 0 m/s

   g = 9.8 m/s²

   s = 0 + 0.5 × 9.8 × 2.2²

  s = 23.72 m

b) impact velocity

      v = √(2gh)

      v = √(2× 9.8 × 23.72)    

      v =  √464.912

      v = 21.56 m/s²  

Answer:

A. d ≈ 23.72 m.

B. v ≈ 21.56 m/s.

Explanation:

d = height

v = final velocity

v0 = Initial velocity

g = gravity

t = time

A. 

Given that the time is 2.2 seconds and the initial velocity is 0, we can replace the values in to the formula.

d = v0 * t + 0.5 * g * t^2

d = 0 * 2.2 + 0.5 * 9.8 * 2.2^2

d = 0 + 23.716

d ≈ 23.72 m.

B.

​

​Sincec the initial speed is 0 m/s,

v^2 = 2 * g * d

v = sqrt(2 * g * d)

v = sqrt(2 * 9.8 * 23.72)

v = sqrt(464.912)

v ≈ 21.56 m/s

Answer:T=[tex]\frac{-26}{r}[/tex]+36

Explanation:

Given Temperature at r=1m is [tex]10^{\circ}C [/tex]

Temperature at r=2m is [tex]28^{\circ}C[/tex]

[tex]\frac{\mathrm{d^2} T}{\mathrm{d} r^2}+\frac{2}{r}\frac{\mathrm{d} T}{\mathrm{d} r}=0[/tex]

Let [tex]\frac{2}{r}\frac{\mathrm{d} T}{\mathrm{d} r}=S[/tex]

[tex]\frac{\mathrm{d^2} T}{\mathrm{d} r^2}=\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}[/tex]

therefore [tex]\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}+\frac{2}{r}[/tex]S=0

[tex]\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}=-\frac{-2S}{r}[/tex]

solving

[tex]r^2[/tex] S=constant

substitute S value

[tex]\frac{\mathrm{d}T}{\mathrm{d}r}[/tex]=[tex]\frac{c}{r^2}[/tex]

Solving it we get

T=[tex]\frac{-c}{r}+c_2[/tex]

Now using given condition

10=[tex]\frac{-c}{1}+c_2[/tex]

28=[tex]\frac{-c}{2}+c_2[/tex]

[tex]c_2[/tex]=36,c=26

putting c values

T=[tex]\frac{-26}{r}[/tex]+36

The expression for the temperature T(r) for the two spheres one inside other is given as,

[tex]T=-\dfrac{26}{r}+36[/tex]

Differential equation is the equation in which there is one or more number of unknown variable exist to find the rate of change of one variable with respect to other.

The radius of the sphere is 1 m and the temperature of the sphere is 10 degree Celsius. The sphere lies inside a concentric sphere with radius 2 m and temperature 28°C.

The differential equation which spheres satisfies is given as,

[tex]\dfrac{d^2T}{dr^2}+\dfrac{2}{r}\dfrac{dT}{dr}=0[/tex]

Let the above equation is equation one,

Suppose S = dT/dr, then

[tex]\dfrac{dS}{dt}=\dfrac{2}{r}\dfrac{d^2T}{dr^2}[/tex]

Put the values in the equation one as,

[tex]\dfrac{2}{r}\dfrac{dS}{dt}+\dfrac{-2}{r}S=0\\\dfrac{2}{r}\dfrac{dS}{dt}=-\dfrac{-2}{r}S\\r^2S=C[/tex]

Here, (C) is the constant value. Now put the value of S as,

[tex]r^2\dfrac{dT}{dr}=C\\\dfrac{dT}{dr}=\dfrac{C}{r^2}\\T=-\dfrac{C}{r}+C_2[/tex]

Put the value of radius as 1 m,

[tex]10=-\dfrac{C}{1}+C_2\\10=-C+C_2[/tex]

Similarly, for the radius 2 meters and temperature 28 degree Celsius,

[tex]28=-\dfrac{C}{2}+C_2[/tex]

On solving above equation, we get,

[tex]C=26\\C_2=36[/tex]

Put the values of constant for the required expression as,

[tex]T=-\dfrac{26}{r}+36[/tex]

Thus, the expression for the temperature T(r) for the two spheres one inside other is given as,

[tex]T=-\dfrac{26}{r}+36[/tex]

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Answer:

a) The object must have constant velocity.

d) The object must have zero acceleration.

Explanation:

We can solve the problem by using Newton's second law, which states that the net force acting on an object is equal to the product between mass and acceleration:

[tex]F = ma[/tex]

where

F is the net force

m is the mass of the object

a is the acceleration

In this problem, the net force on the object is zero:

F = 0

This means that the acceleration of the object is also zero, according to the previous equation:

a = 0

So statement (d) is correct. Moreover, acceleration is defined as the rate of change of velocity:

[tex]a=\frac{\Delta v}{\Delta t}[/tex]

Which means that [tex]\Delta v=0[/tex], so the velocity is constant. Therefore, statement (a) is also correct. The other two statements are false because:

b)The object must be at rest. --> false, the object can be moving at constant velocity, different from zero

c)The object must be at the origin. --> false, since the object can be in motion

If the net force on an object is zero, the object might have constant velocity or zero acceleration. It's not necessary that the object be at rest or at the origin.

The subject of your question is related to the principles of Physics, specifically Newton's First Law of Motion. If the net force on an object is zero, it means that the object is in a state of equilibrium. From the given options:

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Answer:

Explanation:

Arizona Nevada and California. That includes some pretty big cities. Los Angeles, Las Vegas, San Diego to name 3.

A very rigid material—one that stretches or compresses only slightly under large forces—has a large value of Young's modulus.

Answer is A.

A very rigid material—one that stretches or compresses only slightly under large forces—has a large value of Young's modulus. Option a is correct.

A very rigid material—one that stretches or compresses only slightly under large forces—has a large value of Young's modulus. Young's modulus is a measure of the material's stiffness or rigidity. The larger the Young's modulus, the more stress is required to stretch the material to the same extent. For example, steel has a high Young's modulus, while rubber has a low Young's modulus.

Young’s Modulus and Material Rigidity

Young’s modulus, also known as the elastic modulus, is a measure of the stiffness of a material. It quantifies how much a material will deform (i.e., stretch or compress) under a given amount of force. A very rigid material is one that exhibits minimal deformation when subjected to large forces, and this characteristic is directly related to the value of Young’s modulus.

Young’s Modulus and Material Behavior

When a material is subjected to an external force, it deforms in response to that force. This deformation can take the form of stretching (tensile deformation) or compression (compressive deformation). Young’s modulus specifically measures the material’s response to tensile or compressive forces. A high value of Young’s modulus indicates that the material experiences minimal elongation or compression when subjected to these forces, signifying its rigidity.

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Explanation:

It is known that density is the mass present in per unit volume.

Mathematically,         Density = [tex]\frac{mass}{volume}[/tex]

Since, it is given that [tex]d_{1}[/tex] is 1.18 [tex]kg/m^{3}[/tex], [tex]d_{2}[/tex] is 5.30 [tex]kg/m^{3}[/tex], and volume is 2 [tex]m^{3}[/tex].

Therefore, mass of air that has entered will be [tex]m_{2}[/tex] -  [tex]m_{1}[/tex] and it will be calculated as follows.

                       [tex]d_{2} - d_{1}[/tex] = [tex]\frac{m_{2} - m_{1}}{Volume}[/tex]

      [tex]m_{2} - m_{1}[/tex] = [tex](5.30 kg/m^{3} - 1.18 kg/m^{3}) \times 2 m^{3}[/tex]

                                            = 8.24 kg

Thus, we can conclude that mass of air that has entered the tank is 8.24 kg.

Answer:

R = 964.5 m

Explanation:

When plane is moving in vertical loop then at the condition of free fall then force of gravity on the passengers will be balanced by the the pseudo force on them.

In ground frame we can say that normal force on the passengers will become zero

So we have

[tex]mg = \frac{mv^2}{r}[/tex]

[tex]m(9.8) = \frac{mv^2}{R}[/tex]

here we know that

v = 350 km/h = 97.22 m/s

now we have

[tex]9.8 = \frac{97.22^2}{R}[/tex]

[tex]R = 964.5 m[/tex]

Radius of curvature of the flight path must be approximately 966.67 meters to create a condition of apparent zero weight inside the aircraft cabin.

The condition of weightlessness inside an aircraft can be simulated when the centripetal acceleration equals the acceleration due to gravity. According to the formula a = v^2 / r, where a is the centripetal acceleration, v is the velocity, and r is the radius of curvature, we need to rearrange the formula to solve for r: r = v^2 / g. Given that the aircraft is moving at 350 km/h, which is approximately 97.22 m/s (since 1 km/h = 0.27778 m/s), and using g as 9.8 m/s^2, we can calculate the necessary radius of curvature to create a condition of weightlessness. Therefore, the radius of curvature r is:

r = (97.22 m/s)^2 / 9.8 m/s^2

r = 966.67 m

Thus, the radius of curvature of the flight path must be approximately 966.67 meters to create a condition of apparent zero weight inside the aircraft cabin.

Answer:

The correct answer is kinetic energy. :) hope this helps!

Explanation:

Answer:

The ball is in the air for 3.628 sec

Explanation:

We can find speed in the "y" direction when the ball lands with this equation:

[tex]V_{fy} ^{2} = V_{oy} ^{2} + 2*g*(y_{f} - y_{o})[/tex]

Where:   [tex]V_{oy} = 0 m/sec[/tex]   because the ball is thrown horizontally

              [tex]g = - 9.8 m/sec^{2}[/tex]   because gravity is a vector that points down in the "y" direction

              [tex]y_{o} = 64.5 m[/tex]

              [tex]y_{f} = 0 m[/tex]   because the ball lands on the ground

Then:      [tex]V_{fy} ^{2} = (0 m/sec)^{2} + 2(-9.8 m/sec^{2})(0 m - 64.5 m)[/tex]

              [tex]V_{fy} ^{2} = 2(- 9.8 m/sec^{2})(- 64.5 m)[/tex]

              [tex]V_{fy} ^{2} = 1264.2 m^{2}/sec^{2}[/tex]

              [tex]V_{fy} = \sqrt{1264.2 m^{2}/sec^{2}}[/tex]

              [tex]V_{fy} = - 35.5556 m/sec[/tex]   because the ball is falling and that means speed in the "y" direction is a vector that points down

Now we can calculate the time with next equation:

[tex]V_{fy} = V_{oy} + g*t[/tex]

[tex]V_{fy} - V_{oy} = g*t[/tex]

[tex]t = \frac{V_{fy} - V_{oy}}{g}[/tex]

Then:       [tex]t = \frac{-35.5556 m/sec - 0 m/sec}{- 9.8 m/sec^{2}}[/tex]

Finally      t = 3.628 sec

The ball is in the air for approximately 3.63 seconds.

Separate Motions: We can analyze the horizontal and vertical motions of the ball independently since air resistance is neglected.

Horizontal Motion: This is a constant velocity motion because there's no horizontal acceleration (ignoring air resistance).

Vertical Motion: This is a free-fall motion with constant acceleration due to gravity (g ≈ 9.81 m/s²).

Horizontal Distance (x):

Given: x = 103.8 m (horizontal distance from the building base)

Time (t): We need to find the total time (t) the ball spends in the air.

Vertical Motion:

Unknown: Time (t) for the vertical motion (which is the same as the total time in the air)

Known: Vertical distance (y) = -64.5 m (negative since downward) and acceleration due to gravity (g)

Vertical Kinematics Equation: We can use the kinematic equation for constant acceleration to solve for the time (t) for the vertical motion:

y = 1/2 * g * t² (We've plugged in the negative value for y since it's moving downward)

Solve for Time (t) in Vertical Motion:

Rearrange the equation to isolate t: t² = 2 * y / g

Substitute the known values: t² = 2 * (-64.5 m) / (9.81 m/s²) ≈ 13.19 s²

Take the square root of both sides to find t (remember there can be positive and negative time solutions, but we're interested in the positive time for the ball to be in the air).

t ≈ ± 3.63 s (We discard the negative solution)

However, this time represents only the vertical motion.

Relating Horizontal and Vertical Motion: Since the ball is thrown horizontally, the horizontal time (t_horizontal) is the same as the vertical time (t) we just calculated (t_horizontal = t_vertical).

Therefore, the ball is in the air for approximately 3.63 seconds.

Answer:

-99.1 V

Explanation:

For a free electron accelerated from rest, the final kinetic energy of the electron is equal to the change in electric potential energy of the electron:

[tex]K=\Delta U\\\frac{1}{2}mv^2 = q\Delta V[/tex]

where

[tex]m=9.11\cdot 10^{-31}kg[/tex] is the mass of the electron

[tex]v=5.9\cdot 10^6 m/s[/tex] is the final speed

[tex]q=-1.6\cdot 10^{-19} C[/tex] is the charge of the electron

[tex]\Delta V[/tex] is the potential difference

Solving for [tex]\Delta V[/tex], we find

[tex]\Delta V=\frac{mv^2}{2q}=\frac{(9.11\cdot 10^{-31})(5.9\cdot 10^6)^2}{2(-1.6\cdot 10^{-19})}=-99.1 V[/tex]

Final answer:

The spring will stretch approximately 28 inches when the weight is increased to 44lb.

Explanation:

The length that a hanging spring stretches varies directly with the weight placed at the end of the spring. To find out how far the spring will stretch when the weight is increased to 44lb, we can set up a proportion based on the given information:

Proportion:

11lb / 7in. = 44lb / x

Using the cross multiplication method, we can solve for x:

11lb * x = 44lb * 7in.

x = 308in. / 11lb

x ≈ 28in.

Therefore, the spring will stretch approximately 28 inches when the weight is increased to 44lb.

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

[tex]W=K_f - K_i[/tex]

Since the cart was initially at rest, [tex]K_i = 0[/tex], so

[tex]W=K_f = \frac{1}{2}mv^2[/tex] (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, [tex]F_g = 99.5 N[/tex]:

[tex]m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg[/tex]

So solving eq.(1) for v, we find the final speed of the cart:

[tex]v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s[/tex]

2) [tex]2.51\cdot 10^7 J[/tex]

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

[tex]F=4.28 \cdot 10^5 N[/tex]

[tex]d=586 m[/tex]

So the work done is

[tex]W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J[/tex]

3)  [tex]2.51\cdot 10^7 J[/tex]

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

[tex]W=\Delta K = K_f - K_i[/tex]

where

W is the work done

[tex]\Delta K[/tex] is the change in kinetic energy

Therefore, the change in kinetic energy is

[tex]\Delta K = W = 2.51\cdot 10^7 J[/tex]

4) 37.2 m/s

According to the work-energy theorem,

[tex]W=\Delta K = K_f - K_i[/tex]

where

[tex]K_f[/tex] is the final kinetic energy of the train

[tex]K_i = 0[/tex] is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

[tex]W=K_f = \frac{1}{2}mv^2[/tex]

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

[tex]v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s[/tex]

Answer:

(c) no different than on a low-pressure day.

Explanation:

The force acting on the ship when it floats in water is the buoyant force. According to the Archimedes' principle: The magnitude of buoyant force acting on the body of the object is equal to the volume displaced by the object.

Thus, Buoyant forces are a volume phenomenon and is determined by the volume of the fluid displaced.  

Whether it is a high pressure day or a low pressure day, the level of the floating ship is unaffected because the increased or decreased pressure at the all the points of the water and the ship and there will be no change in the volume of the water displaced by the ship.

Answer: C. 8.0 m west

Explanation: The arrows are going 15 m west and 7.0 m east. 7 meters of the west will cancel out because 15-7=8. Subtract the smaller number from the bigger number, which is west minus east. The answer will be 8.0 m west.

The current flowing through the wire at t = 0.7 seconds is approximately 2.67 C/s.

What is the current when t = 0.7 s?

Recall the relationship between current and charge:

Current (I) is the rate of change of charge (Q) over time. Mathematically, it's represented as:

I(t) = dQ(t)/dt

Differentiate the charge function:

Given the function for Q(t) = t³ - 2t² + 4t + 1, we can find the current I(t) by differentiating it with respect to time t:

I(t) = 3t² - 4t + 4

Calculate the current at t = 0.7 s:

Now, plug t = 0.7 into the expression for I(t):

I(0.7) = 3(0.7)² - 4(0.7) + 4

I(0.7) ≈ 2.67 C/s

Therefore, the current flowing through the wire at t = 0.7 seconds is approximately 2.67 C/s.

Answer:

The new pressure inside the container is 394.48 Pa.

Explanation:

Given that,

Pressure = 550.0 kPa

Temperature = 25.0°C

Initial volume = 5.20 L

Final volume = 7.25 L

We need to calculate the new pressure inside the container

Using Boyle's law

PV = constant

[tex]P_{1}V_{1}=P_{2}V_{2}[/tex]

Where, P₁ = Pressure

V₁ = Initial volume

V₂ = Initial volume

Put the volume into the formula

[tex]550\times5.20=P_{2}\times7.25[/tex]

[tex]P_{2}=\dfrac{550\times5.20}{7.25}[/tex]

[tex]P_{2}=394.48\ Pa[/tex]

Hence, The new pressure inside the container is 394.48 Pa.

Final answer:

The new pressure inside the container after expansion to 7.25 L at a constant temperature of 25.0 °C is approximately 544.28 kPa.

Explanation:

The subject of the question at hand involves the behavior of gases and is related to the laws governing gas behavior, specifically Boyle's Law which states that the pressure of a gas is inversely proportional to its volume when temperature and the amount of gas are held constant.

To generate an accurate answer, we must assume that the temperature remains constant at 25.0 °C, as per the question's conditions and use the values given to apply Boyle's Law.

Initial conditions are given as:
P1 = 550.0 kPa
V1 = 5.20 L
The final volume V2 is 7.25 L.
By Boyle's Law (P1V1 = P2V2) we can calculate the new pressure (P2) upon expansion:

P2 = (P1 × V1) / V2
P2 = (550.0 kPa × 5.20 L) / 7.25 L
P2 = 3948 kPa ÷ 7.25 L
P2 ≈ 544.28 kPa

Thus, the new pressure inside the container after expansion to 7.25 L at a constant temperature of 25.0 °C is approximately 544.28 kPa.

Answer:

a) 3.94 rad/s

b) 11.84 rad/s

Explanation:

Given:

Distance, s = 78 cm = 0.78 m

Now the time taken, t

we know

[tex]s = ut +\frac{1}{2}gt^2[/tex]

where,

s  = distance

u = initial speed

g = acceleration due to gravity

since it is a free fall. thus, u = 0

thus, we get

[tex]0.78 = 0\times t +\frac{1}{2}9.8\times t^2[/tex]

or

[tex]t=\sqrt{\frac{2\times 0.78}{9.8}}=0.398 s[/tex]

a) now, the smallest angle will be 1/4 of the revolution so as to fall on the butter side i.e 90° or ([tex]\frac{\pi}{2}[/tex])

also,

[tex]angular\ speed\ (\omega) = \frac{Change\ in\ angle}{time}[/tex]

thus, we have

[tex]angular\ speed\ (\omega) = \frac{\frac{\pi}{2}}{0.398} = 3.94rad/s[/tex]

b)now, the largest angle will be 3/4 of the revolution so as to fall on the butter side i.e 270° or ([tex]\frac{3\pi}{2}[/tex]) contributing to the largest angular speed

[tex]angular\ speed\ (\omega) = \frac{Change\ in\ angle}{time}[/tex]

thus, we have

[tex]angular\ speed\ (\omega) = \frac{\frac{3\pi}{2}}{0.398} =11.84rad/s[/tex]

A. The bullet's speed is [ ( M + m ) / m ] √ ( 2 μ g d )

B. The initial speed of the 9.0 g bullet is about 610 m/s

[tex]\texttt{ }[/tex]

Let's recall Impulse formula as follows:

[tex]\boxed {I = \Sigma F \times t}[/tex]

where:

I = impulse on the object ( kg m/s )

∑F = net force acting on object ( kg m /s² = Newton )

t = elapsed time ( s )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

mass of bullet = m = 9.0 g = 9.0 × 10⁻³ kg

mass of block = M = 12 kg

sliding distance = d = 5.4 cm = 5.4 × 10⁻² m

coefficient of kinetic friction = k = 0.20

Asked:

initial bullet's speed = u₁ = ?

Solution:

Firstly, we will use Conservation of Energy formula to find the speed of the block:

[tex]W = \Delta Ek[/tex]

[tex]fd = \frac{1}{2}(M+m)v^2[/tex]

[tex]\mu N d = \frac{1}{2}(M+m)v^2[/tex]

[tex]\mu (M + m)g d = \frac{1}{2}(M+m)v^2[/tex]

[tex]\mu g d = \frac{1}{2} v^2[/tex]

[tex]\boxed {v = \sqrt{2 \mu g d}}[/tex] → Equation A

[tex]\texttt{ }[/tex]

Next, we will use Conservation of Momentum formula to find the initial speed of the bullet:

[tex]\texttt{Total Momentum Before Collision = Total Momentum After Collision}[/tex]

[tex]m u_1 + M u_2 = ( m + M ) v[/tex]

[tex]m u_1 + M (0) = ( m + M ) v[/tex]

[tex]m u_1 = ( m + M ) v[/tex]

[tex]m u_1 = ( m + M ) \sqrt { 2\mu g d}[/tex] ← Equation A

[tex]\boxed {u_1 = \frac { m + M }{ m } \sqrt { 2\mu g d}}[/tex]

[tex]\texttt{ }[/tex]

[tex]u_1 = \frac { m + M }{ m } \sqrt { 2\mu g d}[/tex]

[tex]u_1 = \frac { 9.0 \times 10^{-3} + 12 }{ 9.0 \times 10^{-3} } \sqrt { 2 \times 0.20 \times 9.80 \times 5.4 \times 10^{-2}}[/tex]

[tex]\boxed{u_1 \approx 610 \texttt{ m/s}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

(a) The expression for the speed of the bullet is [tex]u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1}[/tex]

(b) The speed of the bullet at the given parameters is 613.78 m/s.

The given parameters;

Apply the principle of conservation of linear momentum to determine the initial speed of the bullet;

[tex]m_1 u_1 + m_2 u_2 = V(m_1 + m_2)\\\\m_1 u_1 + 0 = V(m_1 + m_2)\\\\u_1 = \frac{V(m_1 + m_2)}{m_1}[/tex]

Apply the principle of work-energy theorem to determine the speed of the bullet-block system;

[tex]K.E - P.E = W_f\\\\\frac{1}{2} MV^2 - 0 = \mu_k (Mg)d\\\\V^2 = 2\mu_k gd\\\\V = \sqrt{2\mu_k gd[/tex]

The expression for the speed of the bullet is written as;

[tex]u_1 = \frac{V(m_1 + m_2)}{m_1} \\\\u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1}[/tex]

The speed of the bullet at the given parameters is calculated as follows;

[tex]u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1} \\\\u_1 = \frac{(0.009 + 12) \sqrt{2\times 0.2 \times 9.8 \times 0.054} }{0.009} \\\\u_1 = 613.78 \ m/s[/tex]

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Answer:

v = 46.55 m/s

Explanation:

It is given that,

A ball is thrown horizontally from the top of a building 0.10 km high, d = 0.1 km = 100 m

The ball strikes the ground at a point 65 m horizontally away from and below the point of release, h = 65 m

At maximum height, velocity of the ball is 0. So, using the equation of motion as :

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

Here, a = g

[tex]100=0+\dfrac{1}{2}\times 9.8t^2[/tex]

[tex]t=4.51\ s[/tex]

Let [tex]v_x[/tex] is the horizontal velocity of the ball. It is calculated as :

[tex]v_x=\dfrac{65\ m}{4.51\ s}=14.41\ m/s[/tex]

Let [tex]v_y[/tex] is the final speed of the ball in y direction. It can be calculated as :

[tex]v_y^2+u_y^2=2as[/tex]

[tex]u_y=0[/tex]

[tex]v_y^2=2gd[/tex]

[tex]v_y^2=2\times 9.8\times 100[/tex]

[tex]v_y=44.27\ m/s[/tex]

Let v is the speed of the ball just before it strikes the ground. It is given by :

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]

[tex]v=\sqrt{14.41^2+44.27^2}[/tex]

v = 46.55 m/s

So, the speed of the ball just before it strikes the ground is 46.55 m/s. Hence, this is the required solution.

The speed of the ball just before it strikes the ground is equal to 46.55 m/s.

Given the following data:

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex].

To determine the speed of the ball just before it strikes the ground:

First of all, we would determine the time it took the ball to strike the ground by using the formula for maximum height.

[tex]H = \frac{1}{2} gt^2\\\\100 = \frac{1}{2} \times 9.8 \times t^2\\\\200 = 9.8t^2\\\\t^2 = \frac{200}{9.8} \\\\t^2=20.41\\\\t=\sqrt{20.41}[/tex]

Time, t = 4.52 seconds

Next, we would find the horizontal velocity:

[tex]Horizontal\;velocity = \frac{horizontal\;distance}{time} \\\\Horizontal\;velocity = \frac{65}{4.52}[/tex]

Horizontal velocity, V1 = 14.38 m/s

Also, we would find the velocity of the ball in the horizontal direction:

[tex]V_2^2 = U^2 + 2aS\\\\V_2^2 = 0^2 + 2(9.8)(100)\\\\V_2^2 = 1960\\\\V_2 = \sqrt{1960} \\\\V_2 = 44.27 \;m/s[/tex]

Now, we would calculate the speed of the ball just before it strikes the ground by finding the resultant speed:

[tex]V = \sqrt{V_1^2 + V_2^2} \\\\V = \sqrt{14.38^2 + 44.27^2}\\\\V = \sqrt{206.7844‬ + 1959.8329‬}\\\\V =\sqrt{2166.6173}[/tex]

Speed, V = 46.55 m/s

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Answer:

D) 0.33 m/s² to the right

Explanation:

Apply Newton's second law.  Take right to be positive and left to be negative.

∑F = ma

20 N − 5 N = (45 kg) a

a = 0.33 m/s²

The couch accelerates at 0.33 m/s² to the right.

Answer: acceleration = 0.33m/s² to the right.

Explanation: when two forces act in opposite direction along same axis thier resultant direction is always with that force of higher magnitude in this case 20N

Resultant force F = -5N + 20N = 15N to the right.

But

F = Mass * acceleration

acceleration = F/ mass

= 15N/45kg

= 0.33N/kg

Which is also 0.33m/s² to the right.

Final answer:

Using the principle of buoyancy and Archimedes' principle, we can determine the volume of the rock from the change in tension when submerged in water. Then, we use the volume of the rock to calculate the density of the unknown liquid where the tension is 11.4 N.

Explanation:

The problem described involves the principle of buoyancy and Archimedes' principle in particular, which states that the upward buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

When the rock is submerged in water and the tension in the string is 51.9 N, this is the weight of the rock in the air. When the rock is immersed in water and the tension becomes 31.6 N, the apparent weight is reduced due to the buoyant force of water. The difference in tension (51.9 N - 31.6 N = 20.3 N) is the weight of the water displaced, which can be converted into mass (20.3 N / 9.8 m/s2 = 2.07 kg). Knowing that the density of water is 1,000 kg/m3, we can calculate the volume of water displaced and thus the volume of the rock. When the rock is submerged in the unknown liquid and the tension is 11.4 N, a similar calculation for the buoyant force allows us to determine the density of the unknown liquid based on the known volume of the rock.

To solve for the density of the unknown liquid, we first need to determine the volume of the rock using the reduced tension in water (Archimedes' principle). Then we use the reduced tension in the unknown liquid to calculate the density of the unknown liquid. As we're given tensions rather than masses, we translate tension into weight (force) and use the equation for buoyant force (buoyant force = weight in air - apparent weight in fluid) to find the volume displaced. Since the rock displaces an equal volume of fluid, we can find the density of the unknown liquid using the buoyant force equation with the known volume of the rock.

Using the principle of conservation of momentum, the velocity of the 8 kg block, after the 4.3 kg block moves to the right with a speed of 6.7 m/s, is calculated to be -3.6 m/s. The negative sign denotes that the block is moving in the opposite direction to the 4.3 kg block.

The subject of this question is a part of physics known as mechanics, specifically conservation of momentum. The principle of conservation of momentum in a system where no external forces are acting states that the total momentum before an event must be equal to the total momentum after the event. Here, since the system begins with zero total momentum (both blocks initially at rest), it should end with zero total momentum.

In this scenario, after cord burns, the two blocks are free to move. The block with mass 4.3 kg moves to the right. According to the conservation of momentum, the other block will move in the opposite direction (to the left) in order to conserve the total momentum of the system.

We calculate the momentum of the system after the spring is released: Momentum = mass * velocity. For the 4.3 kg block, the momentum would be = 4.3 kg * 6.7 m/s = 28.81 kg*m/s. As the total momentum before the event was zero, the momentum of 8 kg block should be -28.81 kg*m/s (in opposite direction). The velocity of this block can now be calculated by dividing its momentum by its mass, i.e., -28.81 kg*m/s / 8 kg = -3.6 m/s (the negative sign indicates that the velocity is to the left).

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Answer:

Part a)

[tex]W = (10 N)(2 m) = 20 J[/tex]

Part b)

[tex]\Delta K = 0.736 J[/tex]

Part c)

[tex]\Delta U = 13.2 J[/tex]

Part d)

[tex]U_{thermal} = 5.66 J[/tex]

Explanation:

Part a)

Work done by the applied force is given by the formula

[tex]W = F.d[/tex]

here we know that

[tex]F = 10 N[/tex]

[tex]d = 2 m[/tex]

[tex]W = (10 N)(2 m) = 20 J[/tex]

Part b)

As we know that the box was at rest initially and then it is moving with speed 0.80 m/s

so here we can say

[tex]\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/tex]

[tex]\Delta K = \frac{1}{2}(2.3)(0.80)^2 - 0[/tex]

[tex]\Delta K = 0.736 J[/tex]

Part c)

Change in gravitational potential energy is given as

[tex]\Delta U = - W_g[/tex]

[tex]\Delta U = -(-mg sin\theta d)[/tex]

[tex]\Delta U = (2.3)(9.81)(sin17)(2)[/tex]

[tex]\Delta U = 13.2 J[/tex]

Part d)

Now by energy conservation law we can say that

Work done by external agent = change in kinetic energy + change in potential energy + thermal energy lost

so we have

[tex]20 = 13.6 + 0.736 + U_{thermal}[/tex]

[tex]U_{thermal} = 5.66 J[/tex]

Answer:

(a). The work done on the system by force is 20 J.

(b). The change in kinetic energy of the system is 0.736 J.

(c). The change in the gravitational potential energy of the system is 13.18 J

(d). The thermal energy of the system is 6.84 J.

Explanation:

Given that,

Mass of box = 2.3 kg

Force = 10 N

Length = 2.0 m

Angle = 17°

Speed = 0.80 m/s

(a). We need to calculate the work done

Using formula of work done

[tex]W=F\times d[/tex]

Put the value into the formula

[tex]W=10\times2.0[/tex]

[tex]W=20\ J[/tex]

The work done on the system by force is 20 J.

(b). We need to calculate the change in kinetic energy of the system

Using formula of change of kinetic energy

[tex]\Delta K.E=K.E_{f}-K.E_{i}[/tex]

[tex]\Delta K.E=\dfrac{1}{2}mv^2-0[/tex]

Put the value into the formula

[tex]\Delta K.E=\dfrac{1}{2}\times2.3\times(0.80)^2[/tex]

[tex]\Delta K.E=0.736\ J[/tex]

The change in kinetic energy of the system is 0.736 J.

(c). We need to calculate the change in the gravitational potential energy of the system

Using formula of gravitational potential energy

[tex]P.E=mgh\sin\theta[/tex]

Where, h = change in height

Put the value into the formula

[tex]P.E=2.3\times9.8\times2.0\sin17[/tex]

[tex]P.E=13.18\ J[/tex]

The change in the gravitational potential energy of the system is 13.18 J.

(d). We need to calculate the thermal energy of the system

Using formula of thermal energy

Work done=Change in kinetic energy+change in potential energy+change in thermal energy

[tex]\Delta U_{th}=W-\Delta K.E+-Delta P.E[/tex]

Put the value into the formula

[tex]\Delta U_{th}=20-0.736-13.18[/tex]

[tex]\Delta U_{th}=6.084\ J[/tex]

The thermal energy of the system is 6.84 J.

Hence, (a). The work done on the system by force is 20 J.

(b). The change in kinetic energy of the system is 0.736 J.

(c). The change in the gravitational potential energy of the system is 13.18 J

(d). The thermal energy of the system is 6.84 J.

Answer:

100 rad/s

Explanation:

Angular speed = linear speed / radius

ω = v / r

ω = (5.0 m/s) / (0.050 m)

ω = 100 rad/s

The angular speed of the pulley when the string has a linear speed of 5.0 m/s is found using the formula v = rω. So the angular speed is 100 rad/s.

To find the angular speed of the pulley when the string has a linear speed of 5.0 m/s, we can use the relationship between linear speed (v), angular speed (ω), and radius (r) of the pulley. This relationship is given by the formula v = rω, where v is the linear speed, r is the radius of the pulley, and ω is the angular speed we want to find. Given that the radius r is 5.0 cm (which we need to convert to meters by dividing by 100, so r = 0.05 m), and v is 5.0 m/s, we rearrange the formula to solve for ω:

ω = v / r

ω = (5.0 m/s) / 0.05 m

ω = 100 rad/s

Therefore, the angular speed of the pulley is 100 radians per second (rad/s).

Answer:

Determining planetary laws of motion

Explanation

Kepler worked for Tycho Brahe who was recording planetary motion for 20 years. After Brahe's demise Kepler inherited those records of planetary motion. After analyzing those records he put forth the three laws of planetary motion.

Law of Orbits: Every planet in our solar system move in elliptical orbit with sun at one focus It is also called the law of Ellipses

Law of Areas: The imaginary line drawn between the planet and sun will sweep out equal areas in equal period

Law of periods: The square of the time period of the planet is directly proportional to the cube of the semi major axis of the orbit. It is also called the law of Harmonies

Johannes Kepler is known for determining the planetary laws of motion in astronomy. Therefore option 2 is correct.

"determining planetary laws of motion," is the correct answer. Johannes Kepler was a German astronomer and mathematician who made significant contributions to our understanding of the motion of planets.

His three laws of planetary motion, known as Kepler's laws, revolutionized the field of astronomy and laid the foundation for Isaac Newton's later work on universal gravitation.

Kepler's first law, known as the law of ellipses, states that planets orbit the Sun in elliptical paths, with the Sun at one of the foci of the ellipse. This challenged the prevailing notion that planetary orbits were perfect circles.

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Answer: Because its period of rotation is 243.01 days (retrograde)

Explanation:

In comparison, Earth rotates once in about 24 hours with respect to the Sun, but once every 23 hours, 56 minutes, and 4 seconds with respect to other, distant, stars. Earth's rotation is slowing slightly with time, so in the past a day was shorter.

Venus's weak magnetic field can be attributed to its slow retrograde rotation period, lack of a convective liquid metal core, and lack of significant tectonic activity. These conditions are all contrary to those that generally contribute to generating a strong magnetic field.

The weak magnetic field of Venus could be linked to its retrograde rotation period and lack of a convective liquid metal core. Unlike planets with strong magnetic fields, Venus has a retrograde rotation period of 243 days, compared to Earth's 24-hour rotation period. The rotation period plays a key role in generating a planet's magnetic field. A faster rotation aids the creation of a dynamo effect, generating a stronger magnetic field.

Furthermore, the composition of Venus' core could also contribute to its weak magnetic field. A planet's magnetic field is typically caused by the movement of liquid metal within its core, which serves as a good conductor of electricity. It seems that Venus lacks this convective liquid metal core, thereby lacking the necessary conditions for a strong magnetic field.

Additionally, Venus may lack tectonic activity that is known to influence the generation of a magnetic field. While the surface of Venus has been modified by tectonics driven by mantle convection, it does not exhibit the same kind of plate tectonics as Earth.

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