Could I get help on this question please

Answer:

the radius of curvature of the track for this instant is 266 m

Explanation:

Given that;

The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s

velocity of the train at t=12 s is;

[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s

now, we calculate the deceleration of the train

[tex]V_{t=12}[/tex]  = u + at

13.89 = 27.77 + [tex]a_{t}[/tex]12

[tex]a_{t}[/tex] = (13.89 - 27.77) / 12

[tex]a_{t}[/tex] = -13.88 / 12

[tex]a_{t}[/tex] = - 1.1566 m/s²

Now, the velocity of the train at 6 seconds is;

[tex]V_{t=6}[/tex]  = u + at

[tex]V_{t=6}[/tex]  = 27.77 + ( - 1.1566 m/s²)6

[tex]V_{t=6}[/tex]  = 27.77 - 6.9396

[tex]V_{t=6}[/tex]  = 20.83 m/s

The acceleration at t=6 s is;

a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]

a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]

we substitute

2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]

4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²

4 = 1.3377 +  ([tex]a_{n}[/tex])²

([tex]a_{n}[/tex])² = 4 - 1.3377

([tex]a_{n}[/tex])² = 2.6623

[tex]a_{n}[/tex] = √2.6623

[tex]a_{n}[/tex]  = 1.6316 m/s²

Now the radius of curve is;

a = V² / p

[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² /  [tex]a_{n}[/tex]

[tex]p_{t=6}[/tex] = ( 20.83 m/s )² /  1.6316 m/s²

[tex]p_{t=6}[/tex] = 433.8889 / 1.6316

[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m

Therefore;  the radius of curvature of the track for this instant is 266 m

Answer:

McNair graduated as valedictorian of Carver High School in 1967. In 1971, he received a Bachelor of Science degree in engineering physics, magna cu.m laude, from the North Carolina Agricultural and Technical State University in Greensboro, North Carolina.

Answer:

 t = 14.53 s

Explanation:

The speed of a wave is constant and is given by

         v = [tex]\sqrt{ \frac{B}{ \rho} }[/tex]

in this exercise they indicate that we assume the constant velocity, therefore we can use the uniform motion relations

          v = x / t

           t = x / v

in this case the sound pulse leaves the ship and must return so the distance is

          x = 2d

where d is the ocean depth d = 10900m and the speed of sound in seawater is v = 1500 m / s

let's calculate

           t = 2 10900/1500

           t = 14.53 s

Answer:

the required mass flow rate is 49484.37 kg/s

Explanation:

Given the data in the question;

we first determine the relation for mass flow rate of water that passes through the turbine;

so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;

[tex]W_{net}[/tex] = m ( Δ P.E )

so we substitute (gh) for ( Δ P.E );

[tex]W_{net}[/tex] = m (gh)

m = [tex]W_{net}[/tex] / gh

so we substitute our given values into the equation

m = 100 MW / ( 9.81 m/s²) × 206 m

m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m

m = 10 × 10⁷ / 2020.86

m = 49484.37 kg/s

Therefore, the required mass flow rate is 49484.37 kg/s

Answer:

The answer is B .........number 2

Explanation:

Answer:

Explanation:

Let n be number of total number of nucleons ( protons + neutrons )

Total mass inside nucleus =  n x 1.67 x 10⁻²⁷ Kg

volume of nucleus = 4/3 π r³

= 1.33 x 3.14 x (10⁻¹⁵)³ m

= 4.17 x 10⁻⁴⁵ m³

Density = mass / volume

=  n x 1.67 x 10⁻²⁷ / 4.17 x 10⁻⁴⁵

= .4 n x 10¹⁸ kg / m³

or of the order of 10¹⁸ kg/m³

b )

Density of iron = 7900 kg / m³

or of the order of 10⁴ kg / m³

So nucleus of a matter  is about 10¹⁴ times denser than iron .

Answer:

d = 106.41 m

Explanation:

Given that,

Initial speed of the car, u = 60 mph = 26.9 m/s

The deceleration in the car, a = -3.4 m/s²

The average reaction time, t = 0.218 m/s

It finally stops, final velocity, v = 0

We need to find the distance covered by the car as it come to a stop.

Using third equation of motion to find.

[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-26.9^2}{2\times (-3.4)}\\\\d=106.41 m[/tex]

So, the car will cover 106.41 m as it comes to a stop.

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ([tex]Q[/tex]), measured in joules, by the cylindrical resistor is the power multiplied by operation time ([tex]\Delta t[/tex]), measured in hours. That is:

[tex]Q = \dot Q \cdot \Delta t[/tex] (1)

If we know that [tex]\dot Q = 1.2\,W[/tex] and [tex]\Delta t = 86400\,s[/tex], then the amount of heat dissipated by the resistor is:

[tex]Q = (1.2\,W)\cdot (86400\,s)[/tex]

[tex]Q = 103680\,J[/tex]

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ([tex]Q'[/tex]), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ([tex]A[/tex]), measured in square meters:

[tex]Q' = \frac{\dot Q}{A}[/tex] (2)

[tex]Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }[/tex] (3)

Where:

[tex]D[/tex] - Diameter, measured in meters.

[tex]h[/tex] - Length, measured in meters.

If we know that [tex]\dot Q = 1.2\,W[/tex], [tex]D = 4\times 10^{-3}\,m[/tex] and [tex]h = 2\times 10^{-2}\,m[/tex], the heat flux of the resistor is:

[tex]Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }[/tex]

[tex]Q' \approx 4340.589\,\frac{W}{m^{2}}[/tex]

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ([tex]r[/tex]), no unit, is the ratio of the top and bottom surfaces to total surface:

[tex]r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}[/tex] (3)

If we know that [tex]A \approx 2.765\times 10^{-4}\,m^{2}[/tex] and [tex]D = 4\times 10^{-3}\,m[/tex], then the fraction is:

[tex]r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}[/tex]

[tex]r = 0.045[/tex]

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Answer:150M

Explanation:

Answer:

c

Explanation:

it is stored energy because it is built up in said object

Answer:i think it is c

Explanation:

Answer:

Explanation: i think its c to try it

Answer:

energy= MGH

2*9.8*h=180

h=180/19.6

h=9.32 m

The height of the object above the ground would be equal to 9.18 m.

When an object of mass (m) is moved from infinity to a certain point inside the gravitational influence, the amount of work done in displacing it is stored in the form of potential energy and is known as gravitational potential energy.

The mathematical equation for gravitational potential energy can be written as:

Gravitational potential energy = m⋅g⋅h

Where m is the mass, g is the gravitational acceleration and h is the height above the ground.

Given, the mass of the given object, m = 2 Kg

The gravitational potential energy = 180 J

[tex]GPE = m\times g\times h[/tex]

180 = 2 × 9.8 × h

h = 9.18 m

Therefore, the object should be at a height of 9.18 meters in order to have 180 J of gravitational potential energy.

Learn more about gravitational potential energy, here:

https://brainly.com/question/3884855

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Answer:

1.40 m/s^2

Explanation:

Given data

Velocity= 17.4 m/s

time= 12.4 seconds

We want to find the acceleration of the rock

We know that

acceleration = velocity/time

Substitute

acceleration= 17.4/12.4

acceleration=1.40 m/s^2

Hence the acceleration is 1.40 m/s^2

Answer:

10 J.

Explanation:

Given that,

Net force acting on the rake, F = 2 N

Distance moved by the rake, d = 5 m

We need to find the kinetic energy gained by the rake. We know that,

Kinetic energy = work done

So,

K = F×d

K = 2 N × 5 m

K = 10 J

So, 10 J of kinetic energy is gained by the rake.

Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.

How much kinetic energy does the rake gain?

Answer: 10 J

Answer:

I do believe her actions were justified.

Explanation:

Due to the school charging extra fee from her father who makes a modest amount as a teacher. There was sum of money involved that could change how he lived and her.

I do not believe her actions where justified

She had a lot going for her. She could have skipped the hardship of helping grace and pass. She could have easily have gotten a good job with a degree and paid back all the debts owed. Alot of troubles could have been avoided just by doing her own thing.

net force = 30 N

mass = 8.16 kg

acceleration = 3.68 m/s²

Given

80 N force applied

mass of object = 10 kg

Friction force = 50 N

Required

Net force

mass

acceleration

Solution

Net force = force applied(to the right) - friction force(to the left)

Net force = 80 - 50 = 30 N

Gravitational force(downward) : F = mg

m = F : g

m = 80 : 9.8

m = 8.16 kg

a = F net / m

a = 30 / 8.16

a = 3.68 m/s²

Answer:

Explanation:

The electric field inside of a conductor is 0 because the conduction electrons are pushed to the outer edges of the conductor. The surface of the conductor still has charge.

Answer:

Friction can stop or slow down the motion of an object.

Explanation:

The slowing force of friction always acts in the direction opposite to the force causing the motion.

Answer:

13.9 m/s.

Explanation:

Since the vertical velocity of the skydiver is constant at v = 7.0 m/s, we find the time, t it takes him to drop from a height of h = 70 m.

So, distance = velocity time

h = vt

t = h/v = 70 m/7 m/s = 10 s

This is also the time it takes him to move horizontally a distance of d = 120 m to the target.

So, his horizontal velocity is v' = distance/time = d/t = 120m/10 s = 12 m/s.

Since both vertical and horizontal velocities are perpendicular, we add them vectorially to obtain the skydivers total speed, V.

So, V = √(v² + v'²)

= √((7.0 m/s)² + (12.0 m/s)'²)

= √(49 m²/s² + 144 m²/s²)

= √(193 m²/s²)

= 13.9 m/s.

The direction of this velocity is Ф = tan⁻¹(v/v')

= tan⁻¹(7 m/s/12 m/s)

= tan⁻¹(0.5833)

= 30.3°

Answer:

a) v₀ₓ = 62.76 m / s, b)   θ₁ = 17.6º,   θ₂ = 67.0º

Explanation:

We can solve this exercise using the projectile launch ratios

a) Let's find the time it takes for the bullet to reach the water level

       y = y₀ + v_{oy} t - ½ g t²

when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero

       0 = y₀ + 0 - ½ g t²

       t =[tex]\sqrt{2y_o/g}[/tex]

       t = [tex]\sqrt{2 \ 7 /9.8}[/tex]          

       t = 1,195 s

now we can calculate the speed with the horizontal movement

        x = v₀ₓ t

        v₀ₓ = x / t

        v₀ₓ = 75.0 / 1.195

        v₀ₓ = 62.76 m / s

b) if the speed of the bullets is half of that found

         v₀ = 62.76 / 2 = 31.38 m / s

let's write the expressions for the distance

          x = v₀ cos θ t

          y = y₀ + v_{oy} sin θ t - ½ g t²

          t = [tex]\frac{x}{v_o \ cos \theta}[/tex]

we substitute

          [tex]0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2[/tex]

          [tex]0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}[/tex]    

let's use the identified trigonometry

          sec² θ = 1 + tan² θ

         sec θ = 1 / cos θ

we substitute

          [tex]0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)[/tex]

          [tex]\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0[/tex]

we change variable

         tan θ = H

         [tex]\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0[/tex]

we subtitle the values

         [tex]\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0[/tex]

         27.99 H² - 75 H + 20.99 = 0

         H² - 2.679 H + 0.75 = 0

we solve the quadratic equation

         H = [2.679 ± [tex]\sqrt{2.679^2 - 4 0.75}[/tex]] / 2

         H = [2,679 ± 2,044] / 2

         H₁ = 0.3175

         H₂ = 2.3615

now we can find the angles

          H₁ = tan θ₁

          θ₁ = tan⁻¹ H₁

          θ₁ = tan⁻¹ 0.3175

          θ₁ = 17.6º

          θ₂ = 67.0º

for these two angles the bullet hits the boat

C

It is right because I took this and I got this answer correct

Answer:

convection currents

Explanation:

Answer:

a)  [volts] = [N m / C],

b) The lines or surface that has the same potential are called equipotential

c) the equipotential lines must also be perpendicular to the electric field lines

Explanation:

a) find the units of the volt

the electric potential energy is

             V = k q / r

             V = [N m² / C²] C / m

              V = [N m / C]

The electric potential is defined as

             V = E .s

             V = [N / C] [m]

             V = [N m / C] = [volt]

we see that in the two expressions the same result is obtained therefore the volt is

            [volts] = [N m / C]

b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.

c) The electric potential is defined as the gradient of the electric field

             v =  [tex]- \frac{dE}{dx} i^[/tex]

therefore the equipotential lines must also be perpendicular to the electric field lines

(a) The unit of electric potential is volts.

(b) Equipotential lines are lines along which the electric potential is constant.

(c) Equipotential lines are obtained by drawing a line perpendicular to electric field lines.

Electric potential is the work done in moving a unit positive from infinite to a point in the electric filed.

Electric potential = EQ

Electric potential = (V/C) x (C)

Electric potential = Volts.

This is a line along which the electric potential is constant.

Equipotential lines are obtained by drawing a line perpendicular to electric field lines.

Learn more about equipotential lines here: https://brainly.com/question/12074544

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ([tex]\tau[/tex]), measured in Newton-meters, is:

[tex]\tau = I\cdot \alpha[/tex] (1)

Where:

[tex]I[/tex] - Moment of inertia, measured in Newton-meter-square seconds.

[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

[tex]\alpha = \frac{\omega - \omega_{o}}{t}[/tex] (2)

Where:

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

[tex]\omega[/tex] - Final angular speed, measured in radians per second.

[tex]t[/tex] - Time, measured in seconds.

If we know that [tex]\tau = 3\,N\cdot m[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s }[/tex], [tex]\omega = 145.875\,\frac{rad}{s}[/tex] and [tex]t = 4\,s[/tex], then the moment of inertia of the motor is:

[tex]\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}[/tex]

[tex]\alpha = 36.469\,\frac{rad}{s^{2}}[/tex]

[tex]I = \frac{\tau}{\alpha}[/tex]

[tex]I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }[/tex]

[tex]I = 0.0823\,N\cdot m\cdot s^{2}[/tex]

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Answer:

the time for the car to reach the final velocity is 0.56 s.

Explanation:

Given;

acceleration of the car, a = 50 m/s²

final velocity of the car, v = 100 km/h = 27.778 m/s

the initial velocity of the car, u = 0

The time for the car to reach the final velocity is calculated as;

v = u + at

27.778 = 0 + 50t

27.778 =  50t

t = 27.778 / 50

t = 0.56 s

Therefore, the time for the car to reach the final velocity is 0.56 s.

Answer:

- time t taken for car to travel is 64.57 s

- distance travelled between A and B is 1.4887 km

Explanation:

Given the data in the question;

[tex]U_{BC}[/tex] =  83 km/h = ( 83×1000 / 60×60) =  23.0555 m/s

[tex]U_{CD}[/tex] = 41 km/h = ( 41×1000 / 60×60) =  11.3888 m/s

now, we calculate the acceleration;

a =  (  [tex]U_{BC}[/tex] -  [tex]U_{CD}[/tex] ) / t

we substitute

a =  ( 23.0555 -  11.3888 ) / 4.4

a = 11.6667 / 4.4

a = 2.6515 m/s²

Now equation for displacement from BC

[tex]S_{BC}[/tex] =  [tex]U_{BC}[/tex]t + 1/2.at²

we substitute

[tex]S_{BC}[/tex] =  23.0555×4.4 + 1/2×a×(4.4)²

we substitute -2.6515m/s² for a

[tex]S_{BC}[/tex] =  23.0555×4.4 + 1/2×(-2.6515)×(4.4)²

= 101.4442 - 25.6665

[tex]S_{BC}[/tex] = 75.7792 m

Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m

so

[tex]S_{AB}[/tex] +  [tex]S_{BC}[/tex] +  [tex]S_{CD}[/tex]  =  2300 m

we substitute substitute

[tex]S_{AB}[/tex] +  75.7792 m +  [tex]S_{CD}[/tex]  =  2300 m

[tex]S_{AB}[/tex] +  [tex]S_{CD}[/tex] = 2300 - 75.7792

[tex]S_{AB}[/tex] +  [tex]S_{CD}[/tex]  = 2224.2208 m

so we substitute 23.0555t for [tex]S_{AB}[/tex]  and 11.3888t for  [tex]S_{CD}[/tex]  

23.0555t + 11.3888t  = 2224.2208

34.4443t = 2224.2208

t = 2224.2208 / 34.4443

t = 64.57 s

Therefore, time t taken for car to travel is 64.57 s

Distance Between A to B

[tex]S_{AB}[/tex]  = t ×  [tex]U_{AB}[/tex]

we substitute

[tex]S_{AB}[/tex]  = 64.57 s × 23.0555

[tex]S_{AB}[/tex]  = 1488.69 m

[tex]S_{AB}[/tex]  = 1.4887 km

Therefore, distance travelled between A and B is 1.4887 km

Answer:

1. positive acceleration represents an object speeding up; negative acceleration represents an object slowing down

Explanation:

Acceleration is clearly defined as the rate of change of velocity with time. When are body is speeding up as we say, it is accelerating. When a body is coming to rest, it is decelerating.

Positive acceleration occurs when the speed of a moving continues to increase.

Negative acceleration is when the speed of a moving body reduces drastically.

Answer:

b i think so because it makes senes

Answer: When there is another magnet close by.

Explanation: The needle of a compass is itself a magnet, and thus the north pole of the magnet always points north, except when it is near a strong magnet. ... When you take the compass away from the bar magnet, it again points north. So, we can conclude that the north end of a compass is attracted to the south end of a magnet.