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2 years ago

A 2300 Kg car accelerates from rest to 6.00 m/s in 12.00 seconds. What is the net force acting in the car?

2 answers:

6 0

The car's acceleration is (6 m/s) / (12 sec) = 0.5 m/s² .

   Force = (mass) x (acceleration)

   = (2,300 kg) x (0.5 m/s²)

   = 1,150 Newtons

pav-90 [236]2 years ago

4 0

F=ma
a=(v2-v1)/(t2-t1)
a=(6-0)/(12-0)
a=6/12
a= .5 m/s^2
f=2300kg*.5m/s^2
f=1150N
f=1200N if using correct sig figs

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PLEASE HELP ME 45 POINTS

sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)

Where:

$a$ - Net acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

$\Delta y$ - Covered distance, measured in meters.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $\Delta y = 1.5\,m$ , then the time taken by the system is:

$t = \sqrt{\frac{2\cdot \Delta y}{a} }$

$t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }$

$t \approx 0.845\,s$

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

$v = a\cdot t$ (4)

Where $v$ is the final speed of the system, measured in meters per second.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $t \approx 0.845\,s$ , then the final speed of the system is:

$v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)$

$v = 3.551\,\frac{m}{s}$

The final speed of the system is 3.551 meters per second.

8 0

2 years ago

In levelling, the following staff readings were observed involving an inverted staff, A = 2.915 and B = -2.028. What is the rise

salantis [7]

Answer:

The rise from A to B is 0.887

Solution:

As per the question:

The following reading of an inverted staff is given as:

A = 2.915

B = -2.028

Here, for inverted staff, the greater reading shows greater elevation and lesser reading shows lower elevation.

Thus

The rise from A to B is given as:

A - B = 2.915 - 2.028 = 0.887

8 0

2 years ago

Help with these questions please?

egoroff_w [7]

Answer:

5. dispersion

6. 49.8°

Explanation:

5. Dispersion is the name given to the phenomenon of light of different wavelengths being bent differently. A rainbow is the result of light from a point source (the sun) being spread out by wavelength (color), a nice example of dispersion.

___

6. n = 1.31 is the ratio of the sine of the angle of refraction to the sine of the angle of incidence (for light passing to a medium of n = 1). When the angle of refraction is 90°, the angle of incidence is the "critical angle." So, ...

sin(90°)/sin(critical) = 1.31

critical angle = arcsin(1/1.31) ≈ 49.8°

8 0

2 years ago

How are weather satellites and weather stations similar?

alexandr402 [8]

<span>Weather satellites and weather stations are similar, because they both have the same purpose. They are used to help predict future weather as well as as current conditions. The satellites are viewing the weather from a distance at a large scope, but stations are using data they collect on earth to help with the same task.</span>

6 0

3 years ago

Objects that rest have no forces upon them

Pani-rosa [81]

Answer:

false

Explanation:

every object will always have the force of gravity acting upon it.

6 0

2 years ago