Situations during a soccer game in which the ball is determined in or out of play

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0

2 years ago

A wave has frequency of 50 Hz and a wavelength of 10 m. What is the speed of the wave?

kvv77 [185]

Answer:

This is the answer that I got.

Explanation:

Hope it is right.

3 0

2 years ago

A baseball, hit 3 feet above the ground, leaves the bat at an angle of 45° and is caught by an outfielder 3 feet above the groun

bulgar [2K]

Answer:

Initial velocity of ball = 29.92 m/s

Maximum height reached = 22.82 m

Explanation:

Let the initial velocity be b

Considering vertical motion of base ball:-

Initial velocity, u = bsin45 = 0.707b

Acceleration , a = -9.81 m/s²

Displacement, s = 0 m

We have equation of motion s= ut + 0.5 at²

Substituting

s= ut + 0.5 at²

0 = 0.707b x t + 0.5 x -9.81 x t²

t = 0.144b

The time required for the ball to reach out fielder t = 0.144b

Now considering horizontal motion of base ball:-

Initial velocity, u = bcos45 = 0.707b

Acceleration , a = 0 m/s²

Displacement, s = 300 feet = 91.44 m

Time, t = 0.144b

We have equation of motion s= ut + 0.5 at²

Substituting

s= ut + 0.5 at²

91.14 = 0.707b x 0.144b + 0.5 x 0 x (0.144b)²

b = 29.92 m/s

Initial velocity of ball = 29.92 m/s

For finding maximum height reached we need to consider vertical motion of base ball:-

Initial velocity, u = 29.92 x sin45 = 21.16 m/s

Acceleration , a = -9.81 m/s²

Time, t = Half of time of flight = 0.144 x 29.92 x 0.5 = 2.15 s

We have equation of motion s= ut + 0.5 at²

Substituting

s= ut + 0.5 at²

s = 21.16 x 2.15 + 0.5 x -9.81 x 2.15²

s = 22.82 m

Maximum height reached = 22.82 m

8 0

2 years ago

Two blocks of masses 8 kg and 4.3 kg are placed on a horizontal, frictionless surface. A light spring is attached to one of them

oee [108]

Answer:

The velocity of other mass is 3.60 m/s.

Explanation:

Given that,

Mass of first block = 8 kg

mass of second block = 4.3 kg

Speed = 6.7 m/s

We need to calculate the speed of first mass

Using conservation of momentum

$(m_{1}+m_{2})u=m_{1}v_{1}+m_{2}v_{2}$

where, m₁ =mass of first block

m₂ =mass of second block

m₁ =mass of first block

v₂ =speed of second block

Put the value into the formula

$8+4.3\times0=8\times v_{1}+4.3\times6.7$

$v_{1}=\dfrac{-4.3\times6.7}{8}$

$v_{1}=-3.60\ m/s$

Negative sign represent the opposite direction of initial value.

Hence, The velocity of other mass is 3.60 m/s.

3 0

2 years ago

Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli

Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

Explanation:

Given:

Mass of Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = $v_{f}=?$

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

$m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}$

Substituting the values we get

$2\times 14 + 5\times 0 =(2+5)\times v_{f}$

$v_{f}=\dfrac{28}{7}=4\ m/s$

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

8 0

2 years ago