All categories

4 years ago

WHAT IS THE NET FORCE REQUIRED TO GIVE AN AUTOMBILE OF MASS 1600KG AN ACCELERATION OF 4.5M/S2?

Physics

1 answer:

Fed [463]4 years ago

4 0

Answer:

The required net force has a magnitude of 7200 N

Explanation:

Use Newton's 2nd Law to obtain the answer:

$F_{net}= m\,*\,a\\F_{net}=1600 \,*\,4.5 \,= 7200\,\,N$

You might be interested in

The atomic mass of gold is 0.197 kg/mole. how many moles are in 0.566 kg of gold.

iogann1982 [59]

Explanation:

0.566kg *(1mol/0.197 kg)= 2.87 mol gold

note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong

7 0

3 years ago

Read 2 more answers

An unknown material, m1 = 0.45 kg, at a temperature of T1 = 91 degrees C is added to a Dewer (an insulated container) which cont

goldenfox [79]

Answer:

Explanation:

Let the specific heat of material be s

heat lost by material = m₁ s (T 1 - T ) , (T 1 - T ) is fall in temp , m₁ is mass of material

= .45 x s x (91 - 31.4 )

= 26.82 s

Heat gained by water

= m₂ cw (T2 - T )

1.3 x 4186 x ( 31.4 - 23 )

heat lost = heat gained

m₂ cw (T2 - T ) = m₁ s (T 1 - T )

1.3 x 4186 x ( 31.4 - 23 ) = .45 x s x (91 - 31.4 )

45711.12 = 26.82 s

s = 1704.36

7 0

3 years ago

The wheel on an upside-down bicycle moves through 18.0 rad in 5.39 s. What is the wheel’s angular acceleration if its initial an

Elden [556K]

Explanation:

Formula to calculate angular acceleration is as follows.

$\Delta (\theta) = \frac{1}{2} \alpha \Delta t^{2} + \omega_{1} \Delta t$

or, $\alpha = \frac{2(\Delta (\theta) - \omega_{1} \Delta t)}{\Delta t^{2}}$

Putting the given values into the above formula as follows.

$\alpha = \frac{2(\Delta (\theta) - \omega_{1} \Delta t)}{\Delta t^{2}}$

= $\frac{2(18.0 rad - 2.5 rad/s \times 5.30 s)}{(5.39)^{2}}$

= 0.326 $rad/s^{2}$

Thus, we can conclude that the wheel’s angular acceleration if its initial angular speed is 2.5 rad/s is 0.326 $rad/s^{2}$ .

5 0

3 years ago

The strategy implementation tool used to determine what actions are going to be taken, by whom, during what time frame, and with

oee [108]

It's called an action plan

As long as we could organize what actions that are going to be taken, when, where, and what could we achieve by that action, basically we can use anything as an action plan

Action plan is different from a to - do list because an action plan is goal oriented. As long as the goal already accomplished, we don't necessarily have to follow the remaining steps

4 0

4 years ago

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadi

zepelin [54]

Answer:

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.

determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.

a)0.7956kg/s

b)5.437 × 10⁻³m²

Explanation:

The concepts related to the change of mass flow for both entry and exit is applied

The general formula is defined by

$\dot{m}=\rho A V$

Where,

$\dot{m} = mass flow rate\\\rho = Density\\V = Velocity$

values are divided by inlet(1) and outlet(2) by

$\rho_1 = 2.21kg/m^3V_1 = 40m/s$

$A_1 = 90*10^{-4}m^2\\\rho_2 = 0.762kg/m^3\\V_2 = 192m/s$

PART A) Applying the flow equation

$\dot{m} = \rho_1 A_1 V_1\\\dot{m} = (2.21)(90*10^{-4})(40)\\\dot{m} = 0.7956kg/s$

PART B) For the exit area we need to arrange the equation in function of Area, that is

$A_2 = \frac{\dot{m}}{\rho_2 V_2}\\A_2 = \frac{0.7956}{(0.762)(192)}\\A_2 = 5.437*10^{-3}m^2$

7 0

3 years ago