Ask question

Ask question

All categories

2 years ago

10

The field used in the Canadian football League (CFL) has the midfield marker at the 55 yard line.how long is the fiend from goal

line to goal line?

1 answer:

kogti [31]2 years ago

3 0

Answer:

110 yds

Explanation:

Well if 55 yards is 1/2 of the field then 2 x 55 = 110 yards is total field length

You might be interested in

An elevator of massmis initially at rest onthe first floor of a building. It moves upward,and passes the second and third floors

Varvara68 [4.7K]

To solve this problem it is necessary to apply the concepts of Work. Work is understood as the force applied to travel a determined distance, in this case the height. The force in turn can be expressed by Newton's second law as the ratio between mass and gravity, as well

$W = mgh$

Where,

m = mass

h = height

g = Gravitational constant

When it ascends to the second floor it has traveled the energy necessary to climb a height, under this logic, until the 4 floor has traveled 3 times the height h of each of the floors therefore

$h = 3h$

Replacing in our equation we have to

$W = mgh\\W = mg(3h)\\W = 3mgh$

The correct answer is 4.

7 0

3 years ago

In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of 12,000 yards/s. the Col

saw5 [17]

Answer:

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Explanation:

Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ( $a$ ), measured in meters per square second, is estimated by this kinematic formula:

$a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }$ (1)

Where:

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ , $v$ - Initial and final speeds of the spaceship, measured in meters.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v = 10968\,\frac{m}{s}$ and $\Delta s = 212.8\,m$ , then the acceleration experimented by the spaceship is:

$a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}$

$a = 282652.782\,\frac{m}{s^{2}}$

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

5 0

3 years ago

Help please help!!!!!!!!

Sidana [21]

1. Electron

2. Neutron

3. Nucleus

4. Atom

5. Molecule

6 0

2 years ago

You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.

valentinak56 [21]

Answer:

$F=6\times 10^{-7}\ N$

Explanation:

Given:

quantity of point charge, $q=-4\times 10^{-9}\ C$

radial distance from the linear charge, $r=0.09\ m$

linear charge density, $\lambda=3\times 10^{-9}\ C.m^{-1}$

<u>We know that the electric field by the linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.\epsilon_0.r}$

$E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}$

$E=150\ N.C^{-1}$

<u>Now the force on the given charge can be given as:</u>

$F=E.q$

$F=150\times 4\times 10^{-9}$

$F=6\times 10^{-7}\ N$

3 0

3 years ago

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about

nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s$

a) The vertical speed when the player leaves the ground is 4.45 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s$

Time taken to reach the maximum height is 0.45 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s$

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

6 0

3 years ago