A 792 Kg Rollercoaster Car Is On Theinside Of A Loop Of Radius 18.7 M.At The Top Of The Loop, The Car (2024)

According to the Kepler's Third Law of Planetary Motion, if the orbital period of an object around a body is T and the radius of the orbit is R, then, the quotient between the square of the period and the radius elevated to the third power is constant for any object in orbit around the same body:

[tex]\frac{T^2}{R^3}=\text{constant}[/tex]

Let T₁ and R₁ represent the orbital period and the orbital radius of the first planet around Gliese 876, and let T₂ and R₂ represent the same magnitudes for the second planet. Then:

[tex]\frac{T^2_2}{R^3_2}=\frac{T^2_1}{R^3_1}[/tex]

In the problem, R₂ is the unkown that we need to find. Isolate R₂ from the equation:

[tex]\begin{gathered} \Rightarrow R^3_2=\mleft(\frac{T_2}{T_1}\mright)^2R^3_1 \\ \Rightarrow R^{}_2=\mleft(\frac{T_2}{T_1}\mright)^{\frac{2}{3}}R^{}_1 \end{gathered}[/tex]

To find the distance of the second planet to its star, substitute the values for the known variables: T₁=59.9 d, R₁=3.18*10^7 km, T₂=121 d:

[tex]\begin{gathered} R_2=\mleft(\frac{121d}{59.9d}\mright)^{\frac{2}{3}}_{}\times3.18\times10^7\operatorname{km} \\ =5.08\times10^7\operatorname{km} \end{gathered}[/tex]

Therefore, the second planet is 5.08*10^7 km away from its star, Gliese 876.

Given data:

* The initial velocity of the jumper is,

[tex]u=120ms^{-1}[/tex]

* The angle between the initial velocity and the horizontal line is,

[tex]\theta=30^{\circ}[/tex]

Solution:

The horizontal range of the projectile motion by the jumper is,

[tex]H=\frac{u^2\sin (2\theta)}{g}[/tex]

where g is the accleration due to gravity, and H is the horizontal range.

Substituting the known values,

[tex]\begin{gathered} H=\frac{120^2\times\sin (2\times30^{\circ})}{9.8} \\ H=1272.53\text{ m} \end{gathered}[/tex]

Thus, the distance of the jump from its spot is 1272.53 m.

Doorbell:

The doorbell consists of an electromagnet.

The electromagnet is a coil wound on an iron core.

This electromagnet behaves like a magnet when electric current flows through the coil.

This electromagnet attracts the iron strip near it, causing the sound.

The diagram is shown below

Here, iron strip is attracted to the electromagnet.

Given data:

* The weight of the block A is 44 N.

* The weight of the block B is 22 N.

* The coefficient of static friction between A and table is 0.20.

Solution:

(a). The weight of the A and C block is balanced by the weight of the block B.

Thus, the friction force acting on the Block A in terms of the weight of the Block B is,

[tex]W_B=\mu_sW_{AC}[/tex]

This describe the static friction force acting on the Block A ( with block C above it) in contact with the table.

where W_AC is the combined weight of block A and C, W_B is the weight of the block B, and

[tex]\mu_s\text{ is the static friction}[/tex]

Substituting the known values,

[tex]\begin{gathered} 22\text{ = }0.2\times W_{AC} \\ W_{AC}=\frac{22}{0.2} \\ W_{AC}=110\text{ N} \end{gathered}[/tex]

Thus, the minimum weight of the block C in terms of combined weight is,

[tex]\begin{gathered} W_C=W_{AC}-W_A \\ W_C=110-44 \\ W_c=66\text{ N} \end{gathered}[/tex]

Hence, the minimum weight of the block C is 66 N.

Given:

The mass of the boy, M=70 kg

The mass of the snowball, m=3.5 kg

The speed of the snowball before hitting the boy, v=15 m/s

To find:

The speed of the snowball-covered boy after the collision.

Explanation:

From the law of conservation of momentum, the total momentum of a system always remains the same. Thus the total momentum of the boy and the snowball before the collision must be equal to the total momentum of the boy and the snowball after the collision.

Thus,

[tex]\begin{gathered} mv=(m+M)u \\ \Rightarrow u=\frac{mv}{(m+M)} \end{gathered}[/tex]

Where u is the velocity of the snowball-covered boy after the collision.

On substituting the known values,

[tex]\begin{gathered} u=\frac{3.5\times15}{3.5+70} \\ =0.71\text{ m/s} \end{gathered}[/tex]

Final answer:

The speed of the snowball-covered boy after the collision is 0.71 m/s

Thus the correct answer is option B.

ANSWER:

5th option: 80 m/s²

STEP-BY-STEP EXPLANATION:

Given:

Amplitude (A) = 20 cm = 0.2 m

speed in the equilibrium position (v) = 4m/s

To calculate the acceleration, the first thing is to calculate the frequency, using the following formula (maximum speed, i.e. the speed in the equilibrium position):

[tex]\begin{gathered} v=2\pi fA \\ \\ \text{ We replacing} \\ \\ 4=(2)(3.14)(0.2)f \\ \\ f=\frac{4}{(2)(3.14)(0.2)} \\ \\ f=3.185\text{ Hz} \end{gathered}[/tex]

Now, we calculate the value of the maximum acceleration using the following formula:

[tex]\begin{gathered} a=(2\pi f)^2A \\ \\ \text{ We replacing} \\ \\ a=0.2\cdot((2)(3.14)(3.185))^2 \\ \\ a=80.01\cong80\text{ m/s}^2 \end{gathered}[/tex]

Therefore, the maximum acceleration of the system is equal to 80 m/s² so the correct answer is 5th option: 80 m/s²

The mass m needed in order to suspend the leg is 6.73 kg if the leg (with cast) has a mass of 15.0 kg, and its cg is 35.0 cm from the hip joint; the cord holding the sling is 78.0 cm from the hip joint.

τ = r F sin θ

τ = Torque

r = Radius

F = Force

θ = Angle between r and F

θ = 90°

F = m g

τ = r m g

Counter clockwise torque is positive and clockwise torque is taken as negative. So torque due to center of gravity is negative and the torque due to cast is positive.

m1 = 15 kg

r1 = 35 cm = 0.35 m

r2 = 78 cm = 0.78 m

Since the whole system is at rest,

∑ τ = 0

- τ 1 + τ 2 = 0

- r1 m1 g + r2 m2 g = 0

( - 0.35 * 15 * 9.8 ) + ( 0.78 * m2 * 9.8 ) = 0

m2 = 51.45 / 7.64

m2 = 6.73 kg

Therefore, the mass m needed in order to suspend the leg is 6.73 kg

To know more about torque

https://brainly.com/question/28220969

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ANSWER

3) elastic energy

EXPLANATION

When the gymnast reaches the trampolin, both the gymnast and the trampolin move down because of the elasticity of the trampolin, until the trampolin is at its maximum point of which it can stretch. At this point the gravitational energy of the gymnast is zero because of zero height and its kinetic energy is zero too, because of zero energy. Therefore all the energy is stored as elastic energy.

Given data:

* The amount of work done is W = 300 J.

* The time during which the work is done is t = 15 s.

Solution:

The power used in terms of the work done and time taken is,

[tex]P=\frac{W}{t}[/tex]

Substituting the known values,

[tex]\begin{gathered} P=\frac{300}{15} \\ P=20\text{ watts} \end{gathered}[/tex]

Thus, the power used in the given case is 20 watts.

Given,

The mass of the ball, m=0.11 kg

The height of the ball before it was thrown, h₁=0.24 m

The height of the ball after it was caught by Donald, h₂=0.82 m

The gravitational potential energy is the energy stored in an object due to its position above the ground.

The gravitational potential energy of the ball before being thrown is given by,

[tex]E=\text{mgh}_1[/tex]

On substituting the known values,

[tex]\begin{gathered} E=0.11\times9.8\times0.24 \\ =0.26\text{ J} \end{gathered}[/tex]

Thus the gravitational potential energy of the ball relative to the ground, before being thrown is 0.26 J

Answer:

0.073 mg sample will remain

Explanations:

The final amount of a substance after t years given the initial amount and the half life is expressed as:

[tex]A(t)=A_0(\frac{1}{2})^{\frac{t}{x}}[/tex]

where A(t) is the amount after time t

x is the half life

A₀ = initial amount = 75 mg

t = 75 days = 75/365 = 0.2055 years

Half life, x = 7.5 days = 0.02055 years

Substitute these values into the formula given

[tex]\begin{gathered} A=75(0.5)^{\frac{0.2055}{0.02055}} \\ A\text{ = 75(0.5)}^{10} \\ A\text{ = }0.073\text{ mg} \end{gathered}[/tex]

0.073 mg sample will remain

Answer:

steam

Explanation:

Steam can cause worse burns than hot water. This is because when steam touches your skin, it turns back into liquid. When this happens, it releases energy. That energy, along with the heat itself, contributes to how bad the burn is.

Given

Mass of one object, m=4 kg

Velocity of the object, u=29 m/s towards the right

Mass of the other object, m'=11.5 kg

Velocity of the other object, u'=18 m/s towards left

After collision

The object with mass 11.5 kg moves at a speed, v'=4 m/s towards right

To find

The velocity of the object with mass 4 kg after collision

Explanation

Let the direction towards right be positive

By conservation of momentum

[tex]\begin{gathered} mu+m^{\prime}u^{\prime}=mv+m^{\prime}v^{\prime} \\ \Rightarrow4\times29-11.5\times18=4v+11.5\times4 \\ \Rightarrow v=-34.25\text{ m/s} \end{gathered}[/tex]

Conclusion

The velocity of the object is C.-34.25 m/s

In the image you can see it is part of a flow chart, it can be used to describe a process showing its steps, output, inputs, and the desicions that can be taken during the process

Attribut in an ER diagram

The formula for calculating the moment of inertia of a solid sphere is

expressed as

I = 2/5mr^2

where

m is the mass

r is the radius

The formula for calculating the moment of inertia of a hollow sphere is

expressed as

I = 2/3mr^2

By comparing both equatins, we can see that the rotational inertia of the hollow sphere is larger than that of the solid sphere. Recall, inertia is the tendency of a body to resist changes in motion. This means that it is more difficult to speed up the hollow ball or slow down a spinning hollow ball. Acceleration is the rate of change of velocity. Thus, the solid ball would have the greater acceleration

Work (W) done by a force(F) is given by

[tex]\begin{gathered} W=\text{ F.S}\cos\theta\text{ ;}\begin{cases}S={displacement=\text{ 12m}} \\ F={force=\text{ 50N}}\end{cases} \\ \\ \therefore\text{ W= 50}\times12\times1\begin{cases}\theta={angle\text{ between F and S= 0}} \\ cos0\degree={1}\end{cases} \\ \therefore W=\text{ 600 Joule;} \end{gathered}[/tex]

Final answer is 600 Joule

a)

In order to find the initial velocity of the ball, we can use the formulas below:

[tex]\begin{gathered} v_{y0}=v_0\cdot\sin (\theta) \\ v^2_y=v^2_{y0}+2\cdot g\cdot d \end{gathered}[/tex]

At the maximum height, the vertical speed is zero. So, using theta = 30°, d = 2.5 m and vy = 0, we have:

[tex]\begin{gathered} v^2_y=v^2_{y0}+2\cdot g\cdot d \\ 0^2=v^2_{y0^{}}+2\cdot(-9.8)\cdot2.5 \\ v^2_{y0}=49^{} \\ v_{y0}=7 \\ \\ v_{y0}=v_y\cdot\sin (\theta) \\ 7=v_y\cdot\frac{1}{2} \\ v_y=14\text{ m/s} \end{gathered}[/tex]

b)

To find the range, let's calculate the horizontal component of the velocity and the time of flight:

[tex]\begin{gathered} v_x=v\cdot\cos (\theta)_{} \\ v_x=14\cdot\frac{\sqrt[]{3}}{2} \\ v_x=12.124\text{ m/s} \\ \\ v_y=v_{y0}+g\cdot t \\ 0=7-9.8\cdot t \\ t=\frac{7}{9.8} \\ t=0.714 \\ \\ t_f=2t=1.43\text{ s} \\ \\ d_x=v_x\cdot t \\ d_x=12.124\cdot1.43 \\ d_x=17.34\text{ m} \end{gathered}[/tex]

The kinetic energy of an object with mass m that moves with speed v is given by:

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

Plugging the mass and speed given in the equation above we have that:

[tex]\begin{gathered} K=\frac{1}{2}(9.11\times10^{31})(2.74\times10^7)^2 \\ K=3.42\times10^{-16} \end{gathered}[/tex]

Therefore, the kinetic energy of the electron is:

[tex]3.42\times10^{-16}\text{ J}[/tex]

Given data

*The given angular speed is

[tex]\omega=0.76\text{ rad/s}[/tex]

*The radius of the ride is r = 5 m

The formula for the linear speed in meters per second of a person standing on the wall of the ride is given as

[tex]v=\omega r[/tex]

Substitute the known values in the above expression as

[tex]undefined[/tex]

Given:

Length , L = 29.5 cm

Current, I = 19.6 A

B = 0.946

Let's find the following:

(a). Force on the wire if it makes an angle with the magnetic field of 90.0 degrees.

To find the force, apply the formula

[tex]F=BILsin\theta[/tex]

Where:

B is the magnetic field strength = 0.946 T

I is the current = 19.6 A

L is the length of the wire in meters.

Here, the length is in cm, let's convert from cm to m.

We have:

1 cm = 0.01 m

29.5 cm = 29.5 x 0.01 = 0.295 m

Plug in the values and solve for the force, F.

[tex]\begin{gathered} F=0.946*19.6*0.295sin90 \\ \\ F=5.47\text{ N} \end{gathered}[/tex]

Therefore, the force when the angle is 90 degrees is 5.47 N.

• (b). The force on the wire when the angle is 29.6 degrees.

[tex]\begin{gathered} F=0.946*19.6*0.295sin29.6 \\ \\ F=2.7\text{ N} \end{gathered}[/tex]

The force when it makes an angle of 29.6 degrees is 2.70 N.

ANSWER:

(a). 5.47 N

(b). 2.70 N

Components:

First, calculate the cartesian components of each vector, as follow:

[tex]\begin{gathered} A_x=44.0\cdot\cos (28.0)=38.85 \\ A_y=44.0\cdot\sin (28.0)=20.65 \\ B_x=-26.5\cdot\cos (56.0)=-14.82 \\ B_y=26.5\cdot\sin (56.0)=21.97 \\ C_x=0 \\ C_y=-31.0 \end{gathered}[/tex]

Next, consider that the components of the resultant vector R, are given by the sum of the x components and y components of all vectors A, B and C:

[tex]\begin{gathered} R_x=A_x+B_x+C_x=38.5-14.82+0=23.68 \\ R_y=A_y+B_y+C_y=20.65+21.97-31.0=11.62 \end{gathered}[/tex]

Magnitude:

The magnitude is calculated as follow:

[tex]R=\sqrt[]{R^2_x+R^2_y}=\sqrt[]{(23.68)^2+(11.62)^2}=26.38[/tex]

Angle with x axis:

The angle related to the x axis is obtained as follow:

The tangent of the angle related to the x axis is:

[tex]\tan \theta=\frac{R_y}{R_x}[/tex]

which is basically, the quotient between the opposite site and adjacent side of a right triangle formed by the components of the vector.

To obtain the angle you apply tan^-1 to cancel out the tangent, as follow:

[tex]\theta=\tan ^{-1}(\frac{R_y}{R_x})=\tan ^{-1}(\frac{11.62}{23.68})=26.14\degree[/tex]

If the boat moves this way, then the 10km and 7km lines, and the direct distance d form a right triangle. From here we can calculate d using the pythagorean theorem.

Let's determine the various forms of energy used by an old-fashioned alarm clock with springs.

The type of energy the spring has after it is wound can be called the elastic potential energy.

The potential energy can be defined as the energy possessed by an object due to its position due to some zero position.

Therefore, after the spring is wound, it will have the potential energy due to the arrangement/position of the spring.

As the alarm clock runs and performs its various functions, the spring loses energy.

Since the spring is in motion, it loses energy and the potential energy lost becomes kinetic energy.

Kinetic energy can be said to be the energy of an object due to its motion.

Also, due to friction, the some of its potential energy is converted to heat energy.

Therefore, in other words, the energy initially in the spring which is the potential energy, is changed into the kinetic energy and the heat energy.

ANSWER:

The initial energy which is the potential energy is converted into the kinetic energy and heat energy.

Given,

The mass of the ball, m=0.5 kg

The acceleration of the ball, a=50 m/s²

From Newton's second law, the force acting on a body is equal to the product of the mass body and the acceleration produced due to the applied force.

Thus, the applied force is,

[tex]F=ma[/tex]

On substituting the known values,

[tex]\begin{gathered} F=0.5\times50 \\ =25\text{ N} \end{gathered}[/tex]

Thus the applied force is 25 N

Given,

The mass of the skateboard, m₁=2.2 kg

The mass of the jug, m₂=8.7 kg

The speed of the jug after he throws it, v=3.0 m/s

The speed of the boy and skateboard, u=-1.0 m/s

From the law of conservation of momentum, the total momentum of the boy, skateboard, and the jug before the boy throws the jug should be equal to the total momentum of the boy, skateboard, and the jug after he throws it.

As all the objects were at rest before the boy throws the jug, the total momentum initially was equal to zero.

Thus,

[tex](m_1+m_3)u+m_2v=0[/tex]

Where m₃ is the mass of the boy.

On rearranging the equation,

[tex]\begin{gathered} m_1+m_3=\frac{-m_2v}{u} \\ m_3=\frac{-m_2v}{u}-m_1 \end{gathered}[/tex]

On substituting the known values in the above equation,

[tex]\begin{gathered} m_3=\frac{-8.7\times3.0}{-1.0}-2.2 \\ =26.10-2.2 \\ =23.90\text{ kg} \end{gathered}[/tex]

Thus the mass of the boy is 23.90 kg

The average speed is defined as:

[tex]v=\frac{d}{t}[/tex]

Then for the first 30 km it took:

[tex]t=\frac{30}{54}=0.556[/tex]

hours.

The second distance of 50 km took:

[tex]t=\frac{50}{90}=0.556[/tex]

Therefore the average speed is:

[tex]v=\frac{30+50}{0.556+0.556}=72[/tex]

72 km/h.

To convert squared inches to squared centimeters we need to remember that one inch is equal to 2.54 cm.

Now that we know we can do the conversion:

[tex]70in^2\cdot\frac{2.54\text{ cm}}{1\text{ in}}\cdot\frac{2.54\text{ cm}}{1\text{ in}}=\frac{451.612in^2\cdot cm^2}{in^2}=451.612cm^2[/tex]

Therefore 70 squared inches is equal to 451.612 squared cm (this is the exact result). If we use two significant figures then this will be equal to:

[tex]4.5\times10^2cm^2[/tex]

Now, if we want to convert this to squared mm we need to remember that one cm is equal to 10 mm then we would have:

[tex]4.5\times10^2cm^2\cdot\frac{10\text{ mm}}{1\text{ cm}}\cdot\frac{10\text{ mm}}{1\text{ cm}}=4.5\times10^4\operatorname{mm}[/tex]

Therefore we conclude that 70 squares inches is

[tex]4.5\times10^4\operatorname{mm}[/tex]

Given:

Frequency, f = 20 kHz.

Speed of sound in air, v = 331 m/s

Let's find the wavelength for the following:

• (a). When the temperature is 0 degrees Celsius.

To find the wavelength, apply the formula:

[tex]\lambda=\frac{v}{f}[/tex]

Where:

v is the speed

f is the frequency in Hz = 20 x 1000 = 20000 Hz.

To find the speed of sound on a day with 0 C, we have:

[tex]\begin{gathered} v=331*\sqrt{1+\frac{0}{273}} \\ \\ v=331*\sqrt{1} \\ \\ v=331\text{ m/s} \end{gathered}[/tex]

Now, to find the wavelength, input the values into the formula:

[tex]\begin{gathered} \lambda=\frac{v}{f} \\ \\ \lambda=\frac{331}{20000} \\ \\ \lambda=0.01655\text{ m}\approx16.55\text{ mm} \end{gathered}[/tex]

The wavelength when the temperature is 0 C is 16.55 mm.

• (b). Wavelength when the temperature is 40 degrees Celsius.

The speed, v, on a day when the temperature is 40 C will be:

[tex]\begin{gathered} v=331*\sqrt{1+\frac{40}{273}} \\ \\ v=331*\sqrt{1.1465} \\ \\ v=331*1.0708 \\ \\ v=354.42\text{ m/s} \end{gathered}[/tex]

Hence, to find the wavelength, we have:

[tex]\begin{gathered} \lambda=\frac{v}{f} \\ \\ \lambda=\frac{354.42}{20000} \\ \\ \lambda=0.01772\text{ m}\approx17.72\text{ mm} \end{gathered}[/tex]

The wavelength on a day the temperature is 40 C is 17.72 mm.

ANSWER:

• (a). 16.55 mm

• (,b). 17.72 mm.

First, we need to understand what the weight is. Mathematically, the weight is given by:

[tex]\begin{gathered} W=m\cdot g-Fb \\ where: \\ m=mass \\ g=gravity \\ Fb=Force_{\text{ }}of_{\text{ }}bouyancy \end{gathered}[/tex]

In vacuum, there is no force of buoyancy, so in the vacuum the weight reaches its maximum value.

In this sense, the statement is absurd, so we can conclude it is False

Surafce mining can be said to be a form of mining where the rock and the soil covering the mineral deposits are removed.

There are different types of surface mining, which are:

Dregding

Open pit mining

Area strip mining

Contour strip mining

High-wall mining

Mountain top removal

Drift mine is not a type of type of surface mining. Drift mine is a type of underground mining which the entry is just above water level.

Therefore, drift mine is not a type of surface mining.

ANSWER:

Drift mine