A cyclist enters a curve of 30-m radius at a speed of 12 m/s. As the brakes are
applied, speed is decreased at a constant rate of 0.5 m/s2. What are the magnitudes
of the cyclist’s radial and tangential accelerations when his speed is
10 m/s?

Answer:

True

Explanation:

yes we can see that we are helping animals but we create pollution which is very bad

Answer:

option c

Heat is transferred to your hand from a warm cup

conduction is the process of transferring of heat from one material to another when they are in contact

hope it helps

Answer:

A. It has mostly kinetic energy

Explanation:

Kinetic energy refers to movement. Potential energy refers to height. In this case, the big drop just got over. So, when the coaster is at the bottom, it has more kinetic than potential energy . Potential energy is still present but kinetic is more at the bottom.

Answer:

B = 1.37 mT

Explanation:

Given that,

The magnitude of the electric field, E = 480 N/C

The speed of the proton, [tex]v=3.50 \times 10^5\ m/s[/tex]

We need to find the magnitude of the magnetic field. In a velocity selector, the electric field is balanced by the magnetic field. So,

[tex]qE=qvB[/tex]

Where

B is the magnetic field

[tex]B=\dfrac{E}{v}\\\\B=\dfrac{480}{3.5\times 10^5}\\\\B=1.37\times 10^{-3}\ T\\\\or\\\\B =1.37\ mT[/tex]

So, the magnetic field is equal to 1.37 mT.

The phase difference between two points on a progressive wave are out of phase by 0.41 rad is 23°.

An equation is an expression that shows the relationship between two or more numbers and variables.

The two points on a progressive wave are out of phase by 0.41 rad.

Hence, Phase difference = 0.41 rad

But:

Rad to degree = (rad * 180/π)°

Hence:

0.41 rad = (0.41 rad * 180/π) = 23°

The phase difference between two points on a progressive wave are out of phase by 0.41 rad is 23°.

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Answer:

d

Explanation:

Slipping doesn't occur between the belt and cylinder B because the force of static friction is greater than force exerted on cylinder B.

Given the following data:

Mass A = 5 lb to kg = 2.27 kg.

Mass B = 20 lb to kg = 9.02 kg.

Force = 3.6 lb to N = 16.02 Newton.

In order to calculate the angular acceleration of each cylinder, we would take moment about the two cylinders.

For cylinder A:

[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_A\alpha _A = F_A (0.1)\\\\(\frac{m_Ar^2}{2}) \alpha _A = F_A (0.1)\\\\(\frac{2.27 \times 0.1^2}{2}) \alpha _A = F_A (0.1)\\\\0.1F_A=0.01135\alpha _A\\\\F_A=\frac{0.01135\alpha _A}{0.1} \\\\F_A= 0.1135\alpha _A[/tex]

For cylinder B:

[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_B\alpha _B = F_B (0.2)\\\\(\frac{m_Br^2}{2}) \frac{\alpha _A}{2} = F_B (0.2)\\\\(\frac{9.02 \times 0.1^2}{4}) \alpha _A = F_B (0.2)\\\\0.1F_B=0.02255\alpha _A\\\\F_B=\frac{0.02255\alpha _A}{0.2} \\\\F_B= 0.1128\alpha _A[/tex]

For the belt, we have

[tex]\sum F_A =0\\\\P-F_B-F_A=0\\\\16.02-0.1128\alpha _A-0.1135\alpha _A=0\\\\16.02=0.2263\alpha _A\\\\\alpha _A=\frac{16.02}{0.2263} \\\\\alpha _A=70.79 \;rad/s^2[/tex]

Also, we would determine the angular acceleration of cylinder B:

[tex]0.1\alpha _A=0.2\alpha _B\\\\0.1 \times 70.79 = 0.2\alpha _B\\\\7.079= 0.2\alpha _B\\\\\alpha _B=\frac{7.079}{0.2} \\\\\alpha _B=35.40\;rad/s^2[/tex]

Next, we would calculate the forces acting on the cylinders:

[tex]F_A = 0.1135\alpha _A\\\\F_A = 0.1135 \times 70.79\\\\F_A = 8.04 \;Newton[/tex]

[tex]F_B = P-F_A\\\\F_B = 16.02 - 8.04\\\\F_B = 7.98\;Newton[/tex]

Next, we would determine the force of static friction:

[tex]F_s = \mu_s N = \mu_s m_B g\\\\F_s = 0.50 \times 9.02 \times 9.8\\\\F_s=44.198\;Newton[/tex]

From the above calculation, we can deduce that the force of static friction is greater than force exerted on cylinder B. Therefore, slipping doesn't occur between the belt and cylinder B.

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Answer:

The galaxies outside of our own are moving away from us, and the ones that are farthest away are moving the fastest. This means that no matter what galaxy you happen to be in, all the other galaxies are moving away from you

Explanation:

The hypothesis about the movement of the galaxy is that galaxies are moving far from each other continuously.

The milky way galaxy is a galaxy that contains over a hundred billion stars and it also includes our solar system.  Its name describes its appearance when viewed from the earth. All the individual stars in the whole sky are a portion of the Milky Way Galaxy, the term "Milky Way" is because of the band of light.

The astronomer has discovered a new galaxy which means our universe is continuously expanding. This is because the universe encloses everything that exists.

Galaxies are moving in space and since the universe space is continuously expanding so the galaxies continuously move from each other. The farther the galaxy is from the milky way which is an observable part, the faster will be moving the galaxy and the closer the galaxy is to the milky way, the slower will be movement of the galaxy.

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Answer:

5.4 will be the weight in illustrate form

Answer:

a) acceleration

Explanation:

Acceleration is, by definition, the change of an object's velocity.

Answer:

The mass of the block, the distance of the block above the floor, and the time it takes the block to reach the floor, because these quantities can be used to determine the acceleration of the block.

The radius and the mass of the pulley, because these quantities can be used together to determine the rotational inertia of the pulley.

Explanation:

If the motion starts from rest, the initial angular velocity will be zero and the final angular velocity can be determined with the product of angular acceleration and time of motion of the pulley.

Angular velocity is defined as the change in the angular displacement per change in time of motion. This can be expressed mathematically as follows;

[tex]\omega = \frac{\Delta \theta}{\Delta t} = vr[/tex]

where;

In a circular motion that starts from rest and ends with final velocity, the equation is given as;

[tex]\omega_f =\omega_i + \alpha t[/tex]

Where;

Thus, if the motion starts from rest, the initial angular velocity will be zero and the final angular velocity can be determined with the product of angular acceleration and time of motion of the pulley.

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Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

[tex]I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}[/tex] (1)

Where:

[tex]I_{T}[/tex] - Moment of inertia of the turntable, in kilogram-square meters.

[tex]r[/tex] - Distance of the block regarding the center of the turntable, in meters.

[tex]m[/tex] - Mass of the object, in kilograms.

[tex]\omega_{o}[/tex] - Initial angular speed of the turntable, in radians per second.

[tex]\omega_{f}[/tex] - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

[tex]I_{T} = \frac{1}{2}\cdot M\cdot r^{2}[/tex] (2)

Where [tex]M[/tex] is the mass of the turntable, in kilograms.

If we know that [tex]\omega_{o} \approx 7.330\,\frac{rad}{s}[/tex], [tex]M = 1.5\,kg[/tex], [tex]m = 0.54\,kg[/tex] and [tex]r = 0.1\,m[/tex], then the angular speed of the turntable after the event is:

[tex]I_{T} = \frac{1}{2}\cdot M\cdot r^{2}[/tex]

[tex]I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}[/tex]

[tex]I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}[/tex]

[tex]\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}[/tex]

[tex]\omega_{T} = 3.004\,\frac{rad}{s}[/tex] ([tex]28.687\,\frac{rev}{min}[/tex])

The turntable's angular speed after the event is 28.687 revolutions per minute.

Answer:

h

Explanation:

Answer:

the height of the stadium is 4 m

Explanation:

The computation of the height of the stadium is shown below:

but before that total mechanic energy should be determined

E = PE + KE

where

PE = mgh

and, KE = 1 ÷2 mv^2

Now

E = mgh + 1 ÷2 mv^2

= (1.8) (9.8) (2.9) + 1 ÷ 2 (1.8) (4.8)^2

= 71.9J

= 72J

Now the height of the stadium is

TE = mgh

72 = (1.8) × (9.8) × h

So, h = 4 m

Hence, the height of the stadium is 4 m

Answer:

d is trueeeeeeeeeeeeeeeee

Answer:

c

Explanation:

wavelength = speed of light/ frequency

= (3x 10^8 m/s)/(5.0 x 10^14 Hz)

= 6.0 x 10^-7 m

Answer:

[tex]\frac{T_t}{T_c} = 1.32[/tex]

Explanation:

The torque applied on an object can be calculated by the following formula:

[tex]T = Fr[/tex]

where,

T = Torque

F = Applied Force

r = radius of the wheel

For car wheel:

[tex]T_c = Fr_c\\[/tex]

For truck wheel:

[tex]T_t = Fr_t[/tex]

Dividing both:

[tex]\frac{T_t}{T_c} = \frac{Fr_t}{Fr_c}[/tex]

for the same force applied on both wheels:

[tex]\frac{T_t}{T_c} = \frac{r_t}{r_c} \\[/tex]

where,

rt = radius of the truck steering wheel = 0.25 m

rc = radius of the car steering wheel = 0.19 m

Therefore,

[tex]\frac{T_t}{T_c} = \frac{0.25\ m}{0.19\ m} \\[/tex]

[tex]\frac{T_t}{T_c} = 1.32[/tex]

Answer:

a)  v = 3.116 m / s,  b)  μ = 1.65 10⁻²

Explanation:

a) to find the velocity of the set, let's define a system formed by the person and the sled, so that the forces during the collision are internal and the moment is conserved

initial instant. Before the crash

        p₀ = M v₀

final instant. After the crash

        p_f = (M + m) v

the moment is preserved

       M v₀ = (M + m) v

       v = [tex]\frac{M}{M+m} \ v_o[/tex]

let's calculate

      v = [tex]\frac{41.6}{41.6 + 14.6} \ 4.21[/tex]

      v = 3.116 m / s

b) for this part let's use the relationship between work and kinetic energy

        W = ΔK

as the body has its final kinetic energy is zero

the work of the friction forces is

          W = - fr x

the negative sign is because the friction forces always oppose the movement

let's write Newton's second law

Y axis  

        N - W_sled -W_person = 0

        N = mg + M g

        N = (m + M) g

X axis

        fr = ma

the friction force has the expression

        fr = μ N

        fr = μ g (m + M)

we substitute

        - μg (m + M) x = 0- ½ (m + M) v²

          μ = [tex]\frac{1}{2} \ \frac{v^2 }{g \ x }[/tex]

let's calculate

        μ = [tex]\frac{1}{2} \ \frac{3.116^2}{9.8 \ 30.0}[/tex]

        μ = 0.0165

        μ = 1.65 10⁻²

D and E are false.

A, B, and C are true.

Answer:

- Our solar system is made up of the Sun and other objects that orbit the Sun.

- The Sun is the only star in our solar system.

- Dwarf planets have several other bodies in their path orbiting the Sun just as they do.

Explanation:

plato

Answer:

The object is moving at constant speed.

Explanation:

The spaces between the dots are equal.

Answer:

a)   y = 0.35 m,    b) hydrostatic balance

Explanation:

a)  For this fluid mechanics exercise, let's use that the pressure at a given level is the same, let's set a level on the bell shape.

The pressure inside is

             P_interior = P₀ + ρ g h ’

The pressure outside

             P_exterior = Pₐ + ρ g h

as the point is at the same level the pressures are equal

             P_interior = P_exterior

             P₀ + ρ g h ’= Pₐ + ρ g h

             h ’= (Pₐ- P₀)  + ρ g h  

To calculate P₀ they indicate that the pressure increases 10⁵ Pa for every 10 m, we use a direct rule of proportions or rule of three

            P₀ = 10⁵ (40 + h ’) / 10 = 4 10⁵ + h’ 10⁴

the positive sign is because the water inside the hood also increases the air pressure.

we substitute

             (4 10⁵ + h’ 10⁴) + ρ g h’ = Pₐ + ρ g h

             h’ (ρ g + 10⁴) = Pₐ - 4 10⁵ + ρ h h

            h’ (1000 9.8 + 10⁴) = (1 10⁵ -4 10⁵) + 1000 9.8 40

            h' (1.98 10⁴) = -3 105 + 3.92 10⁵

            h’ =  [tex]- \frac{0.92 \ 10^5 }{1.98 \ 10^4 }[/tex]

            h ’= -4.65 m

as the hood is only 5 m high, the free air space is

            Y = 5 - 4.65

             y = 0.35 m

it is very little free space

B) The pressure outside and inside the hood is the same, the water rises inside the hood until the pressures equalize and at this point the force is equal and in the opposite direction, which is why the system is in hydrostatic balance.

Answer:

a)  v = 3,843 m / s, b)  46.7º  North- East

Explanation:

Moment is a vector quantity, so one of the best ways to solve this problem is to solve each component separately.

The system is formed by the two vehicles so that the moment is preserved during the crash

Direction to the East    

initial instant. Before the crash

          p₀ = mₐ vₐ₀

final insttne. After the crash

          p_f = (mₐ + m_b) vₓ

         p₀ = p_f

         mₐ vₐ₀ = (mₐ + m_b) vₓ

         vₓ = [tex]\frac{m_a}{m_a + m_b} \ v_{ao}[/tex]

let's calculate

          vₓ = [tex]\frac{16.7}{16.7 + 29.3} \ 7.26[/tex]

          vₓ = 2,636 m / s

direction north

initial   p₀ = m_b v_{bo}

final     p_f = (mₐ + m_b) v_y

          p₀ = p_f

          m_b v_{bo} = (mₐ + m_b) v_y

          v_y = [tex]\frac{m_b}{m_a+m_b} \ v_{bo}[/tex]

let's calculate

          v_y = [tex]\frac{29.3}{16.7 + 29.3} \ 4.39[/tex]

          v_y = 2.796 m / s

the final speed of the two two vehicles is

          v = (2,636 i ^ + 2,796 j ^) m / s

a) the magnitude of the velocity

let's use the Pythagorean theorem

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

      v = [tex]\sqrt{2.636^2 + 2.796^2}[/tex]

      v = 3,843 m / s

b) let's use trigonometry to find the direction

      tan θ = v_y / vₓ

      θ = tan⁻¹ v_y / vₓ

      θ = tan⁻¹ (2,796 / 2,636)

      θ = 46.7º

This direction is 46.7º  North East

Answer:

Explanation:

I'm going to guess that you want the net force.

These two forces are acting in opposite directions. Therefore the forces are subtracted in effect.

F = F1 - F2

F1 = 22N to the right

F2 = 13N to the left

F = 22 - 13 = 9 N to the right.

Specifying the direction is very important. Forces do have directions and you must specify what that is. Otherwise, the question should be marked incorrect.

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³  

the volume V = 20,000/(1.2×9.8) =  1700 m³