Help Please!!!!!!!!!!!!!!!!!!

Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment

Drupady [299]

Answer:

$u_K=0.862$

Explanation:

The force of friction between the quails feet and the ground is:

$F=m*a$

$F_K=m*a$

$F_K=u_k*m*g$

$u_K*m*g=m*a_c$

$u_K*g=a$

$u_K=\frac{a_c}{g}$

$a_c=\frac{v^2}{r}$

So the coefficient of static is solve

$u_K=\frac{\frac{v^2}{r}}{g}$

$u_K=\frac{v^2}{r*g}=\frac{(2.6m/s)^2}{0.80m*9.8m/s^2}$

$u_K=0.862$

4 0

3 years ago

PLEASE HELP ME

Snowcat [4.5K]

Answer:

Compound.

Explanation:

6 0

4 years ago

Read 2 more answers

What did the asymptote say to the removable discontinuity worksheet answers?

ra1l [238]

“Don't hand that holier than thou line to me” is what the asymptote said to the removable discontinuity.

The distance between the curve and the line where it approaches zero as they tend to infinity is the line in the asymptote of a curve. This is unusual for modern authors but in some sources the requirement that the curve may not cross the line infinitely often is included.

The point that does not fit the rest of the graph or is undefined is called a removable discontinuity. By filling in a single point, the removable discontinuity can be made connected.

6 0

3 years ago

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago

Orlat

Nana76 [90]

The maximum force that the athlete exerts on the bag is equal to 1,500 N and in the opposite direction as the force that the bag exerts on the athlete.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states that action and reaction are equal and opposite.

Fa = -Fb

The force exerted by the athlete on the bag is equal to the force the bag exerted on the athlete but in opposite direction.

Thus, the maximum force that the athlete exerts on the bag is equal to 1,500 newtons and in the opposite direction as the force that the bag exerts on the athlete.

Learn more about force here: brainly.com/question/12970081

#SPJ1

8 0

2 years ago