A star of mass 3.0e30 kg is moving in a circle of radius 1.0e12 metres, with a period of 100 years. This is due to the gravity of a second unseen object. what is the mass of the unseen object in kg

Answers

Answer 1

The mass of the unseen object is 1.21 x 10²⁴ kg.

What is the centripetal acceleration of the star?

The centripetal acceleration of the star is calculated as follows;

a = v²/r

where;

v is the linear speed of the starr is the radius

v = 2πr/T

where;

T is period of the motion = 100 years = 3.154 x 10⁹ s

v = (2π x 1 x 10¹²) / (3.154 x 10⁹)

v = 1,992.1 m/s

a = (1992.1)² / ( 1 x 10¹²)

a = 3.97 x 10⁻⁶ m/s²

The centripetal force of the star is calculated as follows;

F = ma

F = (3 x 10³⁰) x ( 3.97 x 10⁻⁶)

F = 1.19 x 10²⁵ N

The mass of the unseen object is calculated as follows;

F = mg

m = F/g

where;

g is the acceleration due to gravity

m = (1.19 x 10²⁵) / (9.8)

m = 1.21 x 10²⁴ kg

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Related Questions

upon leaving her club, the golf ball moved upward to a height above the surrounding trees. is the ke and pe increasing, decreasing, or staying the same?

Answers

ANSWER

PE increases and KE decreases

EXPLANATION

As described, the golf ball is moving and changing its height, like in the following diagram,

By the law of conservation of energy, the total energy when the ball starts moving and during the whole motion until it stops, must be the same. This total energy is the sum of the potential energy and the kinetic energy.

When the club hits the ball, it gives it a certain amount of kinetic energy but no potential energy. As the ball starts going uphill, the potential energy starts to increase, since it depends on the height of the object. Therefore, to maintain the total energy constant, the kinetic energy must decrease.

Look at Figure 13-17 to answer the question.Will the acceleration of the piano be greater in A or in B? Use Newton's second law ofmotion to explain your answer.

Answers

Answer:

Explanation:

Newton's second law states that the applied force is directly proportional to the rate of change of momentum

momentum = mass x velocity

Rate of change in momentum = (final momentum - initial momentum)/time

Rate of change in momentum = m(v - u)/t

where

v is final velocity

u is initial velocity

t is times

Recall,

Acceleration = (v - u)/t

Thus, the equation becomes

Force = ma

This means that as the force increases, the acceleration increases.

Considering the given scenarios,

First piano is pushed by force from the man

For second piano, there is additional force from the woman and the mass of the piano remains the same. Since there is more force, there is more acceleration. Thus,

Acceleration is greater in B

If the mass of the vase with flowers is 3.2 kg, what is the magnitude of the normal force?

Answers

The magnitude of normal force will be 31.36 N but in the opposite direction of the vase's weight.

We know that forces exist in pairs.

According to Newton's third law of motion,

" Every action has an equal and opposite reaction "

Also,

The second law of motion states, "The magnitude of a force is equal to the product of mass and acceleration acting upon it."

So, keeping in mind the above two laws:

The weight of the vase will be equal to:

W = mass of the vase × acceleration due to gravity

W = m × g

W = 3.2 kg × 9.8 m/s²

W = 31.36 N

So, the magnitude of normal will be the same as the Weight but the direction of Normal will be opposite to the direction of the weight.

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When the reflecting wave flips upside-down on a stretchedstring, which of the following is correct?a The stretched string is with a fixed boundaryb The stretched string is with a free boundaryсThe stretched string may have a fixed boundary ora free.d The given information is not enough.e None of the above is correct.

Answers

We are given a reflecting wave on a string. This can be exemplified in the following diagram:

A 6.5 kg lump of clay is sliding to the right on a fricitonless surface with a speed of 23 m/s. It collides head-on and sticks to a 2 kg metal sphere that is sliding to the left with a speed of -7 m/s. What is the kinetic energy of the combined objects after the collision?

Answers

Answer:

The kinetic energy of the combined objects after the collision = 1768.25 Joules

Explanation:

The mass of the lump of clay, m₁ = 6.5 kg

The speed of the lump of clay, v₁ = 23 m/s

The mass of the metal sphere, m₂ = 2 kg

The speed of the metal sphere, v₂ = -7 m/s

The Kinetic Energy (KE) of the combined objects after collision is calculated as shown below:

[tex]KE=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2^{}[/tex][tex]\begin{gathered} KE=\frac{1}{2}(6.5)(23^2)+\frac{1}{2}(2)(-7)^2 \\ KE=1719.25+49 \\ KE\text{ = }1768.25J \end{gathered}[/tex]

The kinetic energy of the combined objects after the collision = 1768.25 Joules

DQuestion 111 ptsAn object has a mass of 12 kg. Assume the acceleration due to gravity is 10 m/s². If it is lifted to aheight of 20 m, what is its gravitational potential energy?2400 JoulesO 2400 NewtonsO 200 JoulesO 120 Newtons

Answers

Given:

• Mass, m = 12 kg

,

• Acceleration due to gravity, g = 10 m/s²

,

• Height, h = 20 m

Let's find the gravitational potential energy.

To find the gravitational potential energy, apply the formula:

[tex]GPE=m*g*h[/tex]

Where GPE is the gravitational potential energy.

Thus, we have:

[tex]\begin{gathered} GPE=12*10*20 \\ \\ GPE=2400\text{ J} \end{gathered}[/tex]

Therefore, the gravitational potential energy is 2400 Joules.

ANSWERl

Hi can you help me understand how to do this?

Answers

From the information given,

initial velocity of parachute = 198 ft/s

We want to convert 198 ft/s to m/s

Recall,

1 m = 3.3 ft

x m = 198 ft

By crossmultiplying, we have

3.3x = 198

x = 198/3.3

x = 60

Thus,

198 ft/s = 60 m/s

Rate = 60 m/s

To determine the distance that the parachute will fall in 10 seconds, we would apply one of Newton's equations of motion which is expressed as

s = ut + 1/2gt^2

where

s = distance covered

u = initial velocity

t = time

g = acceleration due to gravity and its value is - 9.8m/s^2

From the information given,

t = 10

u = 60

By substituting these values into the formula, we have

s = 60 x 10 - 1/2 x 9.8 x 10^2

s = 600 - 490

s = 110

The distance covered by the parachute in 10s is 110 m

The acceleration of an object is ___proportional to the net force and ___ proportional to its mass? Can you tell me which ones it would be from these options?directly, directlydirectly, inverselyinversely, inverselyinversely, directly

Answers

Answer:

the acceleration of an object is directly proportional to the net force and inversely proportional to its mass.

Explanation:

when the net force on an object goes up, so does the acceleration, meaning one increases as the other increases

the acceleration of an object decreases as the mass increases, therefore this relationship is inversely proportional

What sequence of two displacements moves from (5, 5) m to (- 5, - 5) * m while traveling a distance of exactly 20 meters? How does this distance compare to the single displacement that connects the same starting and ending point?

Answers

The two displacements that move from (5,5) to (-5,-5)

(5,5) → ( 5,-5) [10 units down]

(5,-5) → (-5,5) [ 10 units left ]

The single displacement that connects the 2 points is the hypotenuse of the formed triangle where each side is 10 m long.

Apply Pythagorean theorem:

c2 = 10^2+10^2

c^2 = 100 + 100

c^2 = 200

c =√200

c= 14.14

Compared to the simple displacement (14.14) that connects both points, it is greater.

20m > 14.14 m

A plane is traveling with a velocity of 70 miles/hr with a direction angle of 24 degrees. The wind is blowing at 25 miles/hr with a direction angle of 190 degrees. What is the vertical component of the wind velocity? Round your answer to the nearest whole number.

Answers

Wind velocity:

25 m/h with a direction angle of 190°.

Vertical component:

25 sin 190 = -4.34 m/s = - 4 m/s


Name the instrument which is made on the basis of expansion of heat. ​

Answers

The instrument is the Thermometer

Answer:

It is a thermometer and it helps to see the temperature

27. Scientists have observed an increase in global temperatures over the past 100 years. Which phenomena do scientists believe contributes to the increase in temperatures? A. an increase in undersea volcanic activity B. a decrease in the distance between Earth and the Sun C. an increase in certain gases released during the use of fossil fuels D. a decrease in the amount of water on Earth due to overconsumption

Answers

The answer is letter C) An increase in certain gases released during the use of fossil fuels. Although the others do cause an increase in temperature, their scale cannot be compared to the one caused by fossil fuels.

Two long parallel wires 0.552 meter apart are each carrying 1.75 amperes of current, as shown(a) Find the magnitude and direction of the magnetic field at point A due to the current in the top wire. (b) Find the magnitude and direction of the force per unit length this field exerts on the bottom wire.

Answers

,Given,

Distance between two-wire, d=0.552 m

Current through the wires, I=1.75 A

(a) The magnetic field at a point on the bottom wire due to top wire is given by,

[tex]B_a=\frac{\mu_0I_1}{2\pi d}[/tex]

Where μ₀ is the permeability of the free space.

The direction is given by the right-hand thumb rule. According to this, the direction of the magnetic field produced by top wire will into the plane of the two wires.

On substituting the known values in the above equation,

[tex]B_a=\frac{4\pi\times10^{-7}\times1.75}{2\pi\times0.552}=6.34\times10^{-7}\text{ T}[/tex]

The force per unit length is given by,

[tex]F=I_2\times B_a[/tex]

The direction is given by the right-hand rule. According to this the force is directed towards the top wire.

On substituting the known values in the above equation,

[tex]F=1.75\times6.34\times10^{-7}=1.11\times10^{-6}\text{ N}[/tex]

Therefore the magnetic field acting on the bottom wire due to the current in the top wire is 6.34×10⁻⁷ T and the magnetic force due to this field is 1.11×10⁻⁶ N

In a lightning discharge, 45 C of charge move through a potential difference of 1.0 x108 V in 0.030 s.A. What is the current of the lightning strike?B. How much energy is released by the lightning bolt?

Answers

Given:

Charge, Q = 45 C

Potential difference, V = 1.0 x 10⁸ V

Time, t = 0.030 s

Let's solve for the following:

• (A). What is the current of the lightning strike?

To find the current, apply the formula:

[tex]I=\frac{Q}{t}[/tex]

Where:

I si the current

Q is the charge = 45 C

t is the time = 0.030 s

Thus, we have:

[tex]\begin{gathered} I=\frac{45}{0.030} \\ \\ I=1500\text{ A} \end{gathered}[/tex]

Therefore, the current of the lightning strike is 1500 Amperes.

• (B). How much energy is released by the lightning bolt?

To find the amount of energy released, apply the formula:

[tex]E=V\times Q[/tex]

where:

E is the Energy released

V is the potential difference, V = 1.0 x 10⁸ V

Q is the charge = 45 C

Thus, we have:

[tex]\begin{gathered} E=1.0\times10^8\ast45 \\ \\ E=4.5\times10^9\text{ J} \end{gathered}[/tex]

Therefore, the energy released is 4.5 x 10⁹ Joules.

ANSWER:

(a). 1500 A

(b). 4.5 x 10 J

A computer connected to a motion sensor creates velocity—time graph for a ball rolling down an incline. On the graph, the velocity increases by 0.3 m/s for every tenth of a second increment on the graph. What is the acceleration of the object? A. 0.3 m/s2. B. 1 m/s2. C. 3 m/s2. D. The acceleration cannot be determined from the given information.

Answers

We are asked to determine the acceleration of an object which speed changes by 0.3 m/s every 1/10 seconds. The definition of acceleration is the following:

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

Where:

[tex]\begin{gathered} \Delta v=\text{ change in velocity} \\ \Delta t=\text{ change in time} \end{gathered}[/tex]

Substituting the given values we get:

[tex]a=\frac{0.3\frac{m}{s}}{\frac{1}{10}s}[/tex]

Solving the operations we get:

[tex]a=3\frac{m}{s^2}[/tex]

Therefore, the acceleration is option C.

(20%) Problem 5: Two identical springs, A and B, each with spring constant k = 54.5 N/m, support an object with a weight W = 11.6 N. Each spring makes an angle of 0 = 20.6 degrees to the vertical, as shown in the diagram. Create an expression for the tension in spring A

Answers

The tension in spring A is T = W/(2cosθ)

What is tension?

Tension is the stretching force in a spring.

How to find the expression for the tension in the spring?

Let

T = the tension in the each spring, W =  weight andθ = angle each spring makes with the vertical

Resolving the tension in each spring vertically, so we can have that

for spring A, the tension is Tcosθ and for spring B, the tension is Tcosθ

Now the vertical component of the tension in each equals the weight. So, we have that

Tcosθ + Tcosθ = W

Adding them together, we have that

2Tcosθ = W

Dividing both sides by 2cosθ, so, we can have that

T = W/2cosθ

Thus, the tension in each spring is T = W/(2cosθ)

So, the tension in spring A is T = W/(2cosθ)

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Three vectors are shown in this figure. Their respective moduli are A = 4.00m.B = 3, 20m and C = 2.70mCalculate 2.00 A - B + 1.30 CExpress your answer according toa) Unit vectorsb) The modulus and orientation with respect to the positive part of the x-axis

Answers

Given that,

Modulus of vector A=4.00

The angle made by the vector A with the y axis, θ₁=33.0°

The modulus of vector B=3.20 m

The angle made by the vector B with the x-axis is θ₂=40.0+90.0=130°

The modulus of the vector C=2.70 m

The angle made by the vector C with x-axis θ₃=-90°

The x and y components of the vector can be written as

[tex]\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \end{gathered}[/tex]

Where r is the magnitude (or modulus) of the vector and θ is the angle made by the vector.

Or a vector, in cartesian coordinates, can be written as,

[tex]R=r\cos \theta\hat{\text{i}}+r\sin \theta\hat{j}[/tex]

Therefore, vector A is cartesian coordinates is

[tex]\begin{gathered} \vec{A}=4.00\cos 33^{\circ}\hat{i}+4.00\sin 33.0^{\circ}\hat{j} \\ =3.35\hat{i}+2.18\hat{j} \end{gathered}[/tex]

And the vector B is

[tex]\begin{gathered} \vec{B}=3.20\cos (130^{\circ})\hat{i}+3.20\sin (130^{\circ})\hat{j} \\ =-2.06\hat{i}+2.45\hat{j} \end{gathered}[/tex]

And vector C is given by,

[tex]\begin{gathered} \vec{C}=2.70\cos (-90^{\circ})\hat{i}+2.70\sin (-90^{\circ})\hat{j} \\ =-2.7\hat{j} \end{gathered}[/tex]

The given equation is

[tex]2.00\vec{A}-\vec{B}+1.30\vec{C}[/tex]

Let this represents a vector V

On substituting the known values,

[tex]\begin{gathered} \vec{V}=2.00\vec{A}-\vec{B}+1.30\vec{C} \\ =2.00\times(3.35\hat{i}+2.18\hat{j})-(-2.06\hat{i}+2.45\hat{j)}+1.30(-2.7\hat{j}) \\ =8.76\hat{i}-1.6\hat{j} \end{gathered}[/tex]

(a) This is the representation with the unit vectors, where i and j are the unit vectors along the x-axis and y-axis respectively.

[tex]\vec{V}=8.76\hat{i}-1.6\hat{j}[/tex]

b) The modulus of any vector is the square root of the sum of the squares of its components.

That is, the magnitude of the vector V is

[tex]\begin{gathered} V=\sqrt[]{8.76^2+(-1.6)^2} \\ =8.90\text{ m} \end{gathered}[/tex]

The angle of this vector with the x-axis is given by

[tex]\begin{gathered} \phi=\tan ^{-1}(\frac{-1.6}{8.75}) \\ =-10.36^{\circ}^{} \end{gathered}[/tex]

The negative sign indicates that the vector is below the positive x-axis

Therefore the modulus of the resultant of the above equation is 8.90 m and its angle with the positive x-axis is -10.36°

I got first part correct but dont know how to solve second part: Two new particles with identical positive charge 3 are placed the same 0.0809 m apart. The force between them is measured to be the same as that between the original particles. What is 3 ?

Answers

Answer:

5.92 *10^-6 C

Explanation:

For the two charges q3 the force between them is given by

[tex]F=k\frac{q_3\times q_3}{d^2}[/tex]

Now we know that

F = 48.1 N, d = 0.0809 m, and k = 8.99 *10^9 kg⋅m^3⋅s^−2⋅C^-2; therefore, the above gives

[tex]48.1=(8.99\times10^9)\frac{q_3\times q_3}{(0.0809)^2}[/tex][tex]\Rightarrow48.1=(8.99\times10^9)\frac{(q_3)^2}{(0.0809)^2}[/tex]

Now we solve for q_3.

Dividing both sides by 8.99 * 10^9 gives

[tex]\frac{48.1}{(8.99\times10^9)}=\frac{(q_3)^2}{(0.0809)^2}[/tex]

multiplying both sides by (0.0809)^2 gives

[tex]\frac{48.1}{(8.99\times10^9)}\times\mleft(0.0809\mright)^2=(q_3)^2[/tex]

finally, taking the square root of both sides gives

[tex]\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}=\sqrt{(q_3)^2}[/tex][tex]q_3=\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}[/tex]

Evaluating the right-hand side gives

[tex]\boxed{q_3=_{}5.92\times10^{-6}C\text{.}}[/tex]

Hence, the charge q_3 is 5.92 x 10^-6 C.

The highest frequencies humans can hear isabout 20000 Hz.What is the wavelength of sound in airat this frequency?The speed of sound is310 m/s.

Answers

Given

The highest frequency human can hear is f=20000 Hz

Speed of the sound,v=310 m/s

To find

The wavelength

Explanation

We know,

The wavelength is given by

[tex]\begin{gathered} \lambda=\frac{v}{f} \\ \Rightarrow\lambda=\frac{310}{20000} \\ \Rightarrow\lambda=0.0155\text{ m} \end{gathered}[/tex]

Conclusion

The wavelength is 0.0155 m

A car is traveling at a speed of 62mph. The radius of its tires are 17 inches. What is the angular speed of the tires in rad/min? Round to the nearest whole number.

Answers

To find the angular velocity of the car tires, we can use the angular velocity formula:

[tex]v=rw[/tex]

First, we will need to derive from this equation the equation to find angular velocity as this gives us the speed.

[tex]\begin{gathered} v=rw \\ w=\frac{v}{r} \end{gathered}[/tex]

Now, we need to convert our values from mph to m/s.

1 mph = 0.44704 m/s

62 * 0.44704 = 27.71648 m/s = v

Then, we need to find the radius of the tire in meters.

1 inch = 0.0254 meters

17 * 0.0254 = 0.4318 m = r

Now, we will replace these values with the ones in the equation.

[tex]w=\frac{27.71648}{0.4318}=64.18823529[/tex]

Finally, we can round this to be easier to interpret:

The angular speed of the tire is 64.19 radians/minute, or more compactly: 64.19 rad/min.

a student drops a pebble from the edge of a vertical cliff. the pebble hits the ground 4 s after it was dropped. what is the height of the cliff? a. 20 m b. 40 m c. 60 m d. 80 m

Answers

The object's speed shortly before it lands on the earth is 40 m/s.

What is an example of velocity?

The speed at which something moves in a specific direction is known as its velocity. as the speed of a car driving north on a highway or the pace at which a rocket takes off. Because the velocity vector is scalar, its absolute value magnitude will always equal the motion's speed.

The parameters are as follows: the pebble's time, t = 4 s; the object's velocity right before impact;

The kinematic equation is as follows;

v = in which

v = 0+10 (4)

The object's speed right before impact with the earth is v = 40 m/s2, where g is the acceleration caused by gravity and an is a constant of 10 m/s2. As a result,

the object's final velocity before impact is 40 m/s.

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A 10.0 cm tall object is placed 6.00 cm in front of a curved mirror and produces an image 2.00 cm behind the mirror. What is the focal length of the mirror?0.667 cm1.50 cm-3.00 cm-0.333 cm

Answers

Given data:

The height of object is h₀=10.0 cm.

The object distance is u=6 cm.

The image distance is v=-2.00 cm.(negative because the image is behind the mirror)

The focal length can be calculated by the mirror's formula as,

[tex]\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \frac{1}{f}=\frac{1}{6}+\frac{1}{-2} \\ f=-3.00\text{ cm} \end{gathered}[/tex]

Thus, the focal length of the mirror is -3.00 cm.

An unwary football player collides head-on with a padded goalpost while running at 7.5 m/s and comes to a full stop after compressing the padding and his body by 0.27 m. Take the direction of the player’s initial velocity as positive.1.assuming constant acceleration calculate the his acceleration during the collision in meters per second squared.2 how long does the collision last in seconds.

Answers

Answers:

1. a = -104.16 m/s²

2. t = 0.072

Explanation:

To find the acceleration, we will use the following equation:

[tex]v^2_f=v^2_i+2ax[/tex]

where vf is the final velocity, vi is the initial velocity, a is the acceleration and x is the distance. So, replacing vf by 0 m/s, vi by 7.5 m/s, and x by 0.27m, we get:

[tex]\begin{gathered} 0^2=7.5^2+2a(0.27) \\ 0=56.25+0.54a \end{gathered}[/tex]

Then, solving for a, we get:

[tex]\begin{gathered} 0-56.25=56.25+0.54a-56.25 \\ -56.25=0.54a \\ \frac{-56.25}{0.54}=\frac{0.54a}{0.54} \\ -104.16m/s^2=a \end{gathered}[/tex]

Therefore, the acceleration during the collision is -104.16 m/s²

Then, to calculate how long the collision last, we will use the following equation:

[tex]v_f=v_i+at[/tex]

So, replacing the values and solving for t, we get:

[tex]\begin{gathered} 0=7.5-104.16t \\ 104.15t=7.5 \\ t=\frac{7.5}{104.15}=0.072s \end{gathered}[/tex]

Therefore, the collision last 0.072 seconds

The crazed physic's student's lab partner decides to throw another pumpkin off the
roof with an initial velocity of 19.4 m/s. What is the velocity when the pumpkin
strikes the ground if it takes 3.2 seconds for it to fall?

Please Help :(

Answers

The velocity when the pumpkin strikes the ground if it takes 3.2 seconds for it to fall: 50.78 m/s

From the definition of velocity, we can find the velocity of a falling object is:

v = v₀ + gt

Here

v₀ - 19.4 m/s , t - 3.2  seconds , g - 9.80665 m/[tex]s^2\\[/tex]

v = 19.4 + (9.86 x 3.2)

v = 50.78 m/s

What is velocity?

Velocity and speed describe how quickly or slowly an object is moving. We frequently encounter circumstances when we must determine which of two or more moving objects is going faster. If the two are travelling on the same route in the same direction, it is simple to determine which is quicker. It is challenging to identify who is moving the fastest when their motion is in the other direction. The concept of velocity is useful in these circumstances. Learn about the definition of velocity in this article as well as the distinction between speed and velocity.

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What is the displacement and distance of the car from t=0s to t=10s

Answers

We are given a graph of distance vs time. To determine the displacement we must add up the total distance covered by the car.

We notice from the graph that the distance from t = 0 and t =2 is:

[tex]d_{0-2}=2m[/tex]

From t = 2 and t = 3 there is no change

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Sports biomechanics looks at athletes to better understand how the body:


A. grows.

B. creates energy.

C. moves.

D. resists disease.

Answers

Answer:

The ans is C

Explanation:

In a sporting context, biomechanics examines an athlete in relation to his or her environment and equipment. Forces acting on the body (kinetics) and movements of the body (kinematics) are analysed prior to mapping out an exercise, performance or recovery plan.

If an airplane is running low on fuel, the pilot may decide to dump unneededweight. As the airplane gets lighter, the engines need less fuel to generate thesame amount of acceleration for flight. The pilot has taken advantage ofNewton's law of motion.A. fourthB. thirdO C. firstD. second

Answers

According to Newton's second law we have that the force is equivalent to the product of the mass and the acceleration:

[tex]F=ma[/tex]

This means that if we decrease the mass "m" of the object the required force is decreased also. Since the force is proportional to the required energy and the energy comes from the fuel consumption this means

A baseball player pitches a fastball toward home plate at a speed of 41.0 m/s. The batter swings, connects with the ball of mass 195 g, and hits it so that the ball leaves the bat with a speed of 37.0 m/s. Assume that the ball is moving horizontally just before and just after the collision with the bat.A. What is the impulse delivered to the ball by the bat? Enter a positive value if the impulse is in the direction the bat pushes the ball and enter a negative value if the impulse is in the opposite direction the bat pushes the ball. (kg m/s)B. If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat? (kN)

Answers

Given:

Initial velocity, vi = 41.0 m/s

Mass of ball, m = 195 g = 0.195 kg

Final velocity, vf = 37.0 m/s

Assuming the ball is moving horizontally just before and after collision with the bat, let's solve for the following:

• (A). What is the impulse delivered to the ball by the bat?

To find the impulse, apply the change in momentum formula:

[tex]\Delta p=p_f-p_i[/tex]

Where:

pi is the initial momentum = -mvi

pf is the final momentum = mvf

Thus, we have:

[tex]\begin{gathered} \Delta p=mv_f-(-mv_i) \\ \\ \Delta p=mv_f+mv_i \\ \\ \Delta p=m(v_f+v_i) \\ \\ \Delta p=0.195(37.0+41.0) \\ \\ \Delta p=15.21\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]

Impulse can be said to equal change in momentum.

Therefore, the impulse delivered to the ball by the bat is 15.21 kg.m/s away from the bat.

• (B). If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat?

Apply the formula:

[tex]\text{ Impulse = Force }\ast\text{ time}[/tex]

Rewrite the formula for force:

[tex]\text{ Force=}\frac{impulse}{time}[/tex]

Where:

time = 3.00 m/s

impulse = 15.21 kg.m/s

Hence, we have:

[tex]\begin{gathered} \text{ F=}\frac{15.21}{3} \\ \\ F\text{ = 5.07 kN} \end{gathered}[/tex]

Therefore, the magnitude of the average force exerted on the ball by the bat is 5.07 kN away from the bat.

ANSWER:

(A). 15.21 kg.m/s away from the bat

(B). 5.07 kN.

Problem Try to answer the following questions:(a) What is the maximum height above ground reached by the ball?(b) What are the magnitude and the direction of the velocity of the ball just before it hits the ground? Show Your Problem Solving Steps: Show these below:1) Draw a Sketch2) Choose origin, coordinate direction3) Inventory List – What is known?4) Write the kinematics equation(s) and solution of Part (a):5) Write the kinematics equation(s) and solution of Part (b):Problem 3 A small ball is launched at an angle of 30.0 degrees above the horizontal. It reaches a maximum height of 2.5 m with respect to the launch position. Find (a) the initial velocity of the ball when it’s launched and (b) its range, defined as the horizontal distance traveled until it returns to his original height. As always you can ignore air resistance.(a) Initial velocity [Hints: How is v0 related to vx0 and vy0. How can you use the information given to calculate either or both of the components of the initial velocity?](b) Range [Hints: This problem is very similar to today’s Lab Challenge except that for the challenge the ball will land at a different height.]

Answers

3.

[tex]\begin{gathered} \theta=30^{\circ} \\ y_{\max }=2.5m \end{gathered}[/tex]

a)

[tex]\begin{gathered} y_{\max }=\frac{v^2\sin ^2(\theta)}{g} \\ \end{gathered}[/tex]

Solve for v:

[tex]\begin{gathered} v=\sqrt[]{\frac{y_{\max }\cdot g}{\sin ^2(\theta)}} \\ v=98\cdot\frac{m}{s} \end{gathered}[/tex]

b)

[tex]\begin{gathered} r=\frac{v^2}{9}\sin (2\theta) \\ r=\frac{98^2}{9.8}\cdot\sin (2\cdot30) \\ r=\frac{98^2}{9.8}\sin (60) \\ r=848.7m \end{gathered}[/tex]

Calculate the total capacitance of three capacitors 30µF, 20µF & 12µF connected in parallel across a d.c supply The answer is :Consider that the equivalent capacitance of three capacitors C1, C2 and C3 in parallel is given by:C=C1+C2+C3In this case:C1 = 30µFC2 = 20µFC3 = 12µFReplace the previous values into the formula for C and simplify:C=30μF+20μF+12μF=62μFHence, the total capacitance is 62µFQuestion 5 : Calculate the total charge on the capacitors connected in parallel if the supply voltage is 500V. Sketch a circuit diagram and label this to show how the charges are located

Answers

The circuit diagram is shown below:

From the diagram we notice that the same voltage will flow in every capacitor, this will be helpful later.

We know that this three capacitors are equivalent to a single equivalent capacitor with 62µF capacitance. The charge in this equivalent capacitor is:

[tex]Q_{eq}=(62\times10^{-6})(500)=0.031[/tex]

Now, as we mentioned, the voltage is the same in each capacitor then the charge in each of them is:

[tex]\begin{gathered} Q_1=(30\times10^{-6})(500)=0.015 \\ Q_2=(20\times10^{-6})(500)=0.01 \\ Q_3=(12\times10^{-6})(500)=0.006 \end{gathered}[/tex]

To check if this is correct we need to remember that the charge in the equivalent capacitor is equal to the sum of the charge in each capactior; for this case this conditon is fulfil; therefore we conclude that:

• The charge in the first capacitor is 0.015 C

,

• The charge in the second capacitor is 0.01 C

,

• The charge in the third capacitor is 0.006 C

The diagram with the labels is shown below:

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