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3 years ago

A 1,200 kg car is accelerated at 3.7 m/s2. What force was needed to produce this acceleration?

Physics

2 answers:

Mashutka [201]3 years ago

4 0

The answer would be A

kondaur [170]3 years ago

3 0

Answer:

A. 4,440 N

Explanation:

A Force applied on an object with mass 'm' can change its velocity in a unit time. This change is called acceleration.

If a force 'F' is applied on a body of mass 'm', it produces an acceleration 'a' in the body.

Force is give as :

F = ma

In our case,

m = 1200 kg a = 3.7 m/s²

F = 1200 x 3.7

F = 4440 kgm/s²

F = 4440 N

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A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.

irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ( $\tau$ ), measured in Newton-meters, is:

$\tau = I\cdot \alpha$ (1)

Where:

$I$ - Moment of inertia, measured in Newton-meter-square seconds.

$\alpha$ - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

$\alpha = \frac{\omega - \omega_{o}}{t}$ (2)

Where:

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\omega$ - Final angular speed, measured in radians per second.

$t$ - Time, measured in seconds.

If we know that $\tau = 3\,N\cdot m$ , $\omega_{o} = 0\,\frac{rad}{s }$ , $\omega = 145.875\,\frac{rad}{s}$ and $t = 4\,s$ , then the moment of inertia of the motor is:

$\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}$

$\alpha = 36.469\,\frac{rad}{s^{2}}$

$I = \frac{\tau}{\alpha}$

$I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }$

$I = 0.0823\,N\cdot m\cdot s^{2}$

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

5 0

2 years ago

Select the correct answer. What is tan for the given triangle? 0.60 1.67 1.33 5.00 0.20

GuDViN [60]

The answer is 1.33 i hope this helps you

8 0

3 years ago

Read 2 more answers

Two forces act on a moving object that has a mass of 27 kg. One has a magnitude of 12 N and points due south, while the other ha

levacccp [35]

0.77 m/s2 directed 35° south of west

net force = (-17,-12)

net force = mass * acceleration

(-17,-12) = 27 * (x-acceleration,y-acceleration)

(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)

angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.

magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)

7 0

2 years ago

4. A meter has a resistance of 100 Ω and gives a full scale deflection when it carries a current of 25 μA. (a) What resistor, Rx

frez [133]

Answer:

A=50mΩ

B≅50mΩ

Explanation:

A) To answer this question we have to use the Current Divider Rule. that rule says:

$Ix=.\frac{Req}{Rx} *Itotal$ (1)

Itotal represents the new maximun current, 50mA, Ix is the current going through the 100 ohms resistor, and Req. is the equivalent resitor.

We now have a set of two resistor in parallel, so:

$Req.=\frac{1}{\frac{1}{R1}+\frac{1}{R2} }$ (2)

where R1 is the resitor we have to calculate, and R2 is the 100 ohms resistor (25 uA).

substituting and rearranging (2)

$Req.=\frac{ 100*R1}{R1+100}$ (3)

Now substituting (3) in (1).

$25*10^{-6} =\frac{\frac{ 100*R1}{R1+100}}{100} *50*10^{-3}$

solving this, The value of R1 is: 50mΩ

This value of R1 will guaranty that the ammeter full reflection willl be at 50mA.

Given that R2 (100ohm) it too much bigger than 50mΩ, the equivalent resistor will tend to 50mΩ

If you substitude this values on (2) Req. will be 49.97 mΩ.

7 0

3 years ago

How to calculate the magnitude of the horizontal component of a vector<br>

trapecia [35]

Taking a look at the image in the attachment, we discover that we can calculate the magnitude of the horizontal components using our knowledge of trigonometry. Since we are comparing the resultant and the horizontal component, the equation connecting them is

$cos \theta = \frac{V_{x}}{V}$ , where $V_{x}$ is the horizontal component, and $V$ is the resultant vector. Now we have to make

$V_{x} = Vcos \theta$ , and this is how we calculate the magnitude of the horizontal component.

$V_{x} = Vcos \theta$

8 0

2 years ago