A car initially traveling at 15 m/s speeds up at a constant rate of 2.0 m/s2 for 3 seconds. The velocity of the car at the end of the 3 second interval is _________ m/s.

Answer:

The second option - the papers are acted upon by an unbalanced force from the fan.

Explanation:

Answer:

mass and on the net force acting on it. ... Tap again to see term ... Newton's second law of motion states that an object's acceleration depends on its ... You hit a ping-pong ball & a tennis ball with a tennis rack

Answer:

the magnitude F of the total friction force is 2456.7 lb

Explanation:

Given the data in the question;

maximum speed = 35 mi/hr = ( 35×5280 / 60×60) = 51.3333 ft/s

diameter = 200ft

radius = 200/2 = 100 ft

First we calculate the normal component of the acceleration;

[tex]a_{n}[/tex] = v² / p

where v is the velocity of the car( 51.3333 ft/s)

p is the radius of the curvature( 100 ft)

so we substitute

[tex]a_{n}[/tex] = (51.3333 ft/s)² / 100ft

[tex]a_{n}[/tex] = (2635.1076 ft²/s²) / 100ft

[tex]a_{n}[/tex] = 26.35 ft/s²

we convert Feet Per Second Squared (ft/s²) to Standard Gravity (g)

1 ft/s² = 0.0310809502 g

[tex]a_{n}[/tex] = 26.35 ft/s² × 0.0310809502 g

[tex]a_{n}[/tex] =  0.8189g

Now consider the dynamic equilibrium of forces in the Normal Direction;

∑[tex]F_{n}[/tex] = m[tex]a_{n}[/tex]

F = m[tex]a_{n}[/tex]

we know that mass of the car is 3000-lb =  3000lb([tex]\frac{1}{g}slug[/tex]/1 lb)

so

we substitute

F =  3000lb([tex]\frac{1}{g}slug[/tex]/1 lb)  × 0.8189g

F = 2456.7 lb

Therefore; the magnitude F of the total friction force is 2456.7 lb

Answer:

first is Atoms

4) is True

Answer: 4

Explanation: because my pet monkey said it was

The potential energy of the football with mass 2.5 kg which is lifted up to the top of a cliff is 4410 Joules.

Potential energy is the stored energy which depends upon the relative position of the various parts of a system of objects. Potential energy is the product of mass of the object, acceleration due to gravity, and the height. The SI unit of potential energy is Joule (J).

PE = m × g × h

PE = Potential energy,

m = mass of the object,

g = acceleration due to gravity,

h = height

PE = 2.5 × 9.8 × 180

PE = 4410 Joules

Therefore, the potential energy of the football is 4410 Joules.

Learn more about Potential energy here:

https://brainly.com/question/24284560

#SPJ1

Answer:

Explanation:

Pressure due to water column as deep as 10994 meters can be given by the following expression

Pressure = h d g , where h is depth of water , d is density of water and g is acceleration due to gravity .

Pressure = 10994 x 10³ x 9.8

= 10.77 x 10⁷ N / m²

Pressure will act on curved surface of the spherical shell , the effective surface area will be π R² where R is radius of the surface .

Effective surface = 3.14 x 0.1²

= .0314 m²

Total force = pressure due to water column x effective surface

= 10.77 x 10⁷  x .0314 N.

= 33.82 x 10⁵ N .

Answer:  a) weight on Earth = mass of the object and gravity n the Earth. = 65*10 = 650 kg.

Explanation:

An astronaut has a mass of 65 kg on Earth where the gravitational field strength is 10 N kg A calculate the astronaut's weight on Earth

          hope this helps :)

Answer:

650

Explanation:

use the equation

weight = gm

Answer: light from the object

Explanation:

When light is reflected off an object like a lamp or a door is travels in a straight line but in a new direction so if the light enters our eyes we will see the object because our eyes can detect light

Answer:

38.09 m

Explanation:

We'll begin by calculating the distance travelled by the car during the reaction time. This can be obtained as follow:

Reaction time (tᵣ) = 0.404 s

Initial velocity (u) = 22.8 m/s,

Distance travelled during the reaction time (sᵣ) =?

sᵣ = utᵣ

sᵣ = 22.8 × 0.404

sᵣ = 9.21 m

Next, we shall determine the distance travelled by the car when the brake was applied. This can be obtained as follow:

Initial velocity (u) = 22.8 m/s

Acceleration (a) = –9 m/s² (since the car is decelerating)

Final velocity (v) = 0 m/s

Distance travelled when the brake was applied (s₆) =?

v² = u² + 2as₆

0² = 22.8² + (2 × –9 × s₆)

0 = 519.84 – 18s₆

Collect like terms

0 – 519.84 = –18s₆

–519.84 = –18s₆

Divide both side by –18

s₆ = –519.84 / –18

s₆ = 28.88 m

Finally, we shall determine the stopping distance of the car, as measured from the point where the driver first notices the red light. This can be obtained as follow:

Distance travelled during the reaction time (sᵣ) = 9.21 m

Distance travelled when the brake was applied (s₆) = 28.88 m

Stopping distance =?

Stopping distance = sᵣ + s₆

Stopping distance = 9.21 + 28.88

Stopping distance = 38.09 m

Answer:

Explanation:

Horizontal component of initial velocity of throw = 20 cos 53 = 12 m /s

Vertical component = 20 sin 53 = 15.97 m /s

Distance to be travelled horizontally = 6 m .

time taken by ball to travel this distance = 6 / 12 = 0.5 s

vertical displacement during this period can be calculated as follows .

Initial vertical velocity = 15.97 m /s

time of travel = .5 s

acceleration = - 9.8 m /s²

s = ut - 1/2 g t²

= 15.97 x .5 - .5 x 9.8 x 0.5²

= 7.985 - 1.225

= 6.76 m

Goal post is 6 m high , so ball will cross the goal post .76 m or 76 cm above cross bar  .

b ) vertical component of ball when it crosses the goal post . Let it be v .

v = u - gt

Applying this formula for vertical movement ,

v = 15.97 - 9.8 x .5

= 15.97 - 4.9

= 11.07 m /s .

Answer:

Charge at rest only produces electric field. Moving charge produces both electric field and magnetic field.

plz follow me

Answer:

     l = 0.548 m

Explanation:

For this exercise we compensate by finding the speed of the car

         p = m v

         v = p / m

         v = 0.58 / 0.2

         v = 2.9 m / s

this is how fast you get to the ramp, let's use conservation of energy

starting point. Lowest point

         Em₀ = K = ½ m v²

final point. Point where it stops on the ramp

         [tex]Em_{f}[/tex] = U = m g h

  mechanical energy is conserved

          Em₀ = Em_{f}

          ½ m v² = m g h

           h = [tex]\frac{m v^2}{2 g}[/tex]

let's calculate

          h = [tex]\frac{0.2 \ 2.9^2}{2 \ 9.8}[/tex]

          h = 0.0858 m

to find the distance that e travels on the ramp let's use trigonometry, we look for the angle

          tan θ = y / x

          tan θ = 12/75 = 0.16

          θ = tan⁻¹ 0.16

          θ = 9º

therefore

           sin 9 = h / l

           l = h / sin 9

           l = 0.0858 / sin 9

           l = 0.548 m

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:

it 1.5 meters

Explanation:

if u could put the option number it will be cool and hope it help and if it doesnt am really sorry ;)

Answer:

2m/s

Explanation:

v=f×wavelength

v=2×1

=2m/s

Answer:

a. 2500 cm³.

b. 2.5 litres.

Explanation:

Given the following data:

Density = 0.8g/cm³

Mass = 2000g

To find the volume of the petrol;

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

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

Making volume the subject of formula, we have;

[tex]Volume = \frac{mass}{density}[/tex]

Substituting into the equation, we have:

[tex]Volume = \frac{2000}{0.8}[/tex]

Volume = 2500 cm³

a. The volume of the petrol in the can in cubic centimeters (cm³) is 2000.

b. The volume of the petrol in the can in litres;

1000 cm³ = 1 litre

2500 cm³ = x litres

Cross-multiplying, we have;

1000x = 2500

x = 2500/1000

x = 2.5 litres.

Therefore, the volume of the petrol in the can in litres is 2.5.

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:

Explanation:

We shall first calculate the velocity at height h = 575 m .

acceleration a = 2.2 m /s²

v² = u² + 2 a s

u is initial velocity , v is final velocity , s is height achieved

v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

v = 0 , u = 50.3 m /s

H = ?

0 = 50.3² - 2 x 9.8 H

H = 129.08 m

Total height attained by rocket

= 575 + 129.08

= 704.08 m .

Answer:

The ribbon will move to the right.

Explanation:

To know the the correct answer to the question, we shall determine the net force and direction. This can be obtained as follow:

Force to the right (Fᵣ) = 120 N

Force to the left (Fₗ) = 80 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 120 – 80

Fₙ = 40 N to the right.

From the calculation made above, the net force is 40 N to the right. Thus, the ribbon will move to the right.

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:

θ = 225 rad

Explanation:

given data

angle = 25 rad

to find out

angular velocity after 3t?

solution

let angular acceleration α in t

θ = ω × t + 0.5 × α × t²        ........................1

here ω  = 0 (initial velocity )

so put this value here

25 = 0 + 0.5 × α × t²             ..........................2

α = 25 ÷ (0.5 t²)

α = 50 ÷ t²                            .........................3

now here we take in 3t

θ = ω × 3t + 0.5 × α × (3t)²

for ω  = 0

θ = 0 + 0.5 × α × 9t²

now put value in eq 2

so

θ = (0.5) ×  (50 ÷ t²) ×  (3t)²

θ = 25 × 9

θ = 225 rad

Answer:

ionized particles from the sun.

* interactions in radiation belts.

* the friction of the planet in the solar wind

q = +9 10⁵ C

Explanation:

Due to being made up of matter, the planet Earth has a series of positive and negative charges, in general these charges should be balanced and the net charge of the planet should be zero, but there are several phenomena that introduce unbalanced charges, for example:

* ionized particles from the sun.

* interactions in radiation belts.

* the friction of the planet in the solar wind

This creates that the planet has a net electrical load

We can roughly calculate the charge of the planet

             E = k q / r²

             q = E r² / k

let's calculate

             q = 200 (6.37 10⁶)²/9 10⁹

              q = +9 10⁵ C

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = [tex]\frac{2}{5}[/tex] m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ ([tex]\frac{2}{5}[/tex] m r²) ([tex]\frac{v}{r}[/tex])²

         m g h = ½ m v² (1 + [tex]\frac{2}{5}[/tex])

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = [tex]\frac{7}{5}[/tex] ([tex]\frac{1}{2}[/tex] m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

therefore the block reaches higher than the sphere

         [tex]\frac{y_{sphere}} {y_bolck} = 5/7[/tex]

Answer:

84.1 m

Explanation:

Given :

The distance from the ship to submarine :

20 + y

Using Pythagoras :

Tan θ = opposite / Adjacent

Tan θ = 20 / 12

12 tan θ = 20

θ = tan^-1(20/12)

20

θ = 59.036°

The angle phi;

n1sinθ1 = n2sin θ

Sin 59.036 = 1.33 * sin phi

Sin phi = sinsin(59.04) ÷1.33

0.8574907 = 1.33 * sin phi

Sin phi = 0.8574907 / 1.33

Sin phi = 0.6447298

phi = sin(0.6447298

Phi = 40.15°

From Pythagoras :

y = 76tan40.15°

y = 76 * 0.8435707

y = 64.11

20 + y

20 + 64.11 = 84.11

Answer:

a=1m/s^2

Explanation:

1÷1÷1=1m/s^2

Answer:

Please find attached pdf

Explanation: