Two ice skaters approach each other at right angles. Skater A has a mass of 68.2 kg and travels in the x direction at 2.48 m/s. Skater B has a mass of 34.4 kg and is moving in the y direction at 1.18 m/s. They collide and cling together. Find the final speed of the couple. Answer in units of m/s.

Answer:

D

Explanation:

Work = Distance x Mass

work done = 100,000 FT LB

Work is done whenever a force moves something over a distance or The work done by a force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement.

Work done = force * displacement

given :

force = 500 pound

displacement = 200 feet

work done = 500 * 200 = 100,000 FT LB

correct option is d. 100,000 FT LB

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Answer:

I answered Number 4 (Solids and Elasticity)

Explanation:

solids and elasticity

Answer:i think it is c

Explanation:

Answer:

Explanation: i think its c to try it

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:

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

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Answer:

the required height is 0.2449 m only

Explanation:

Given the data in the question;

Initial height = 2m

so speed of the puck before hitting the ground will be;

u² = 2gh

Initial speed u_ball = √2gh

u_ball = √( 2 × 9.8 × 2 )

u_ball =  √39.2

u_ball = 6.26 m/s

given that; FOR THE FROZEN PUCK, coefficient of restitution = 0.35 only

R = - (v_ball - v_ground / u_ball - u_ ground)

so

0.35 = - (v_ball - 0 / 6.26 - 0)

0.35 = -v_ball / - 6.26

-v_ball = 0.35 × (- 6.26)

-v_ball = -2.191 m/s

v_ball = 2.191 m/s

to get the height;

v² = 2gh

h = v² / 2g

we substitute

h = (2.191)² / 2×9.8

h = 4.800481 / 19.6

h = 0.2449 m

Therefore, the required height is 0.2449 m only

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:

A star's brightness also depends on its proximity to us. The more distant an object is, the dimmer it appears.

Answer:

The variables that can be measured in joules are

B. Energy

D. Work

Hope it will help :)

Answer:150M

Explanation:

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:

There is no correlation that can be made between the predator and prey populations.

Explanation:

Ermrmrm.

Answer:

124.51 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 49.4 m/s

Final velocity (v) = 0 m/s (at maximum height)

Maximum height (h) =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

The maximum height to which the cannon ball attained before falling back can be obtained as illustrated below:

v² = u² – 2gh ( since the ball is going against gravity)

0² = 49.4² – (2 × 9.8 × h)

0 = 2440.36 – 19.6h

Collect like terms

0 – 2440.36 = –19.6h

–2440.36 = –19.6h

Divide both side by –19.6

h = –2440.36 / –19.6

h = 124.51 m

Therefore, maximum height to which the cannon ball attained before falling back is 124.51 m

Answer:

yes, we can tell if there will be rain by the swelling in rain clouds, we can also see if a tornado is forming based on the look of the wall cloud. we can use satellites to predict the amount of precipitation or wind.

Explanation:

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Answer:

2400°C

Explanation:

Volume of tank = 0.4 m^3

steam pressure = 6 Mpa

Steam temperature = 40°C

Initial pressure of tank = 0.1 MPa

Final pressure of Tank = 6 Mpa

Calculate the temperature in the tank when the Pressure in Tank = 6Mpa

since the volume of the Tank is constant = 0.4 m^3

we will apply Gay-Lussac's Law

= [tex]\frac{T1}{P1} = \frac{T2}{P2}[/tex]  ------ ( 1 )

T1 =  40°c

P1 = 0.1 MPa

P2 = 6 Mpa

T2 = ?

From equation 1 above

T2 = ( T1 * P2 ) / P1

     = ( 40 * 6 ) / 0.1

     = 2400°C

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 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:

  μ = 0.66,    therefore if it compiesy with the regulations

Explanation:

Let's solve this exercise in part, let's start by finding with kinematics the acceleration of man

           y = v₀ t + ½ a t²

as it starts from rest the initial velocity is zero

          y = ½ a t²

          a = [tex]\frac{2y}{t^2}[/tex]

          a =\frac{2 \ 4.38}{1.21^2}  

          a = 6.46 m / s²

Now let's use Newton's second law,

Axis y

         N- W = 0

         N = W

         N = m g

X axis

on this axis the man exerts a backward force and by the law of action and reaction the floor exerts a forward force of the same magnitude, this forward force is the friction force.

         fr = m a

the friction force has an expression

          fr = my N

let's substitute

      μ mg = m a

      μ = a / g

let's calculate

     μ = 6.46 / 9.8

     μ = 0.66

therefore if you comply with the regulations

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:

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:

About 1 meter

Explanation:

Base on your image, when the period is 2 (horizontal variable), the length of the pendulum is about 1 meter.

Answer: 1 Newton

Explanation:

"Every action has an equal and opposite reaction."

Please mark as Brainliest if it is correct.

Force is an action-reaction principle. It stated that the force always exists in a pair. The reaction force of the wall on the swimmer will be 1N.

Newton's third law of motion state that every action has an equal and opposite reaction. It is an action-reaction principle. It stated that the force always exists in a pair.

Both occur in an action-reaction form. Hence for every action, there is an equal and opposite reaction.

[tex]\rm F_{action}=F_{reaction} \\\\ \rm F_{action= 1N[/tex]

[tex]\rm F_{reaction} = 1N[/tex]

Hence the reaction force of the wall on the swimmer will be 1N.

To learn more about Newton's third law refer to the link;

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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:

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:

first is Atoms

4) is True

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:

By paying close attention to the formula for average kinetic energy, we can see that by increasing the mass by a proportional amount will lead to an increase in the total average kinetic energy. There is a direct relationship being observed between the values.