All categories

2 years ago

A. Describe one meal that you typically order when eating out.

Physics

2 answers:

SIZIF [17.4K]2 years ago

5 0

He is right because he just know what he is doing

marta [7]2 years ago

3 0

A. “I usually get 10 piece nugget meal with medium fries at McDonald’s with a McFlurry

B. I could replace the 10 piece nuggets with a grilled chicken wrap and replace the medium fries with a small fries then replace the mcflurry with either water or unsweetened tea.

You might be interested in

Only 2 questions plz answer.

dimaraw [331]

1).B
2.)C
i think the reason was because of the pangea theory that continents where once one big continent and over time it began to split

8 0

2 years ago

Read 2 more answers

A 2.0 kg 2.0kg2, point, 0, start text, k, g, end text cart moving right at 5.0 m s 5.0 s m 5, point, 0, start fraction, st

butalik [34]

Yes así le tienes que hacer para aprender

7 0

2 years ago

Read 2 more answers

An object with mass M is attached to the

zysi [14]

Answer:

W = 5/4 Mgℓ

Explanation:

Sum of forces:

∑F = ma

T − Mg = M (g/4)

T = 5/4 Mg

Work = force × distance

W = Tℓ

W = 5/4 Mgℓ

8 0

2 years ago

Read 2 more answers

PLS HELP THIS IS WORTH 180 PTS!!(PLUS I WILL MARK BRAINIEST IF SOMEONE ANSWERS CORRECTY FIRST)

vazorg [7]

$\qquad\qquad\huge\underline{{\sf Answer}}$

The correct choice is ~ A

Bikes don't require gasoline for their functioning, so it causes less air pollution.

7 0

1 year ago

Read 2 more answers

A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

7 0

2 years ago