what is the largest and smallest possible resultant force of two force with magnitude of 41N and 14N

1. Quais motivos fazem com que o Croquet Golf seja a variação da

KiRa [710]

Probablemente porque es un deporte tranquilo y relajante si de eso estás hablando

Espero que esto ayude

3 0

3 years ago

From what height must an oxygen molecule fall in a vacuum so that its kinetic energy at the bottom equals the average energy of

alexandr1967 [171]

Answer:

The value is $h = 11930 \ m$

Explanation:

From the question we are told that

The temperature is $T = 300 \ K$

Generally the root mean square speed of the oxygen molecules is mathematically represented as

$v = \sqrt{\frac{3 * R * T }{M} } = \sqrt{ 2 * g * h }$

Here R is the gas constant with a value $R = 8.314 \ J\cdot K^{-1} \cdot \ mol^{-1}$

M is the molar mass of oxygen molecule with value $M = 0.032 \ kg /mol$

So

$\frac{3 * 8.314 * 300 }{0.032} = 2 * 9.8 * h$

=> $h = 11930 \ m$

4 0

3 years ago

?/1 Jorge traveled 5 miles north to school. He then traveled 3 miles west to the store. Then he left the store and traveled 5 mi

aleksklad [387]

Answer:

He's 3 miles west of school.

Explanation:

He went 5 miles up and 5 miles down which means that he really didn't go up or down. In between that, he went 3 miles west so if the 5 milers don't count, this puts him at 3 miles west of school.

4 0

2 years ago

The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular

nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

6 0

3 years ago

According to recent typical test data, a Ford Focus travels 0.240 mi in 19.3 s, starting from rest. The same car, when braking f

Anit [1.1K]

Answer:

Explanation:

a )

While breaking initial velocity u = 62.5 mph

= 62.5 x 1760 x 3 / (60 x 60 ) ft /s

= 91.66 ft / s

distance trvelled s = 150 ft

v² = u² - 2as

0 = 91.66² - 2 a x 150

a = - 28 ft / s²

b ) While accelerating initial velocity u = 0

distance travelled s = .24 mi

time = 19.3 s

s = ut + 1/2 at²

s is distance travelled in time t with acceleration a ,

.24 = 0 + 1/2 a x 19.3²

a = .001288 mi/s²

= 2.06 m /s²

c )

If distance travelled s = .25 mi

final velocity v = ? a = .001288 mi / s²

v² = u² + 2as

= 0 + 2 x .001288 x .25

= .000644

v = .025 mi / s

= .0025 x 60 x 60 mi / h

= 91.35 mph .

d ) initial velocity u = 59 mph

= 86.53 ft / s

final velocity = 0

acceleration = - 28 ft /s²

v = u - at

0 = 86.53 - 28 t

t = 3 sec approx .

4 0

3 years ago

what is the largest and smallest possible resultant force of two force with magnitude of 41N and 14N​

what is the largest and smallest possible resultant force of two force with magnitude of 41N and 14N