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OleMash [197]
3 years ago
9

you push with an 18-N horizontal force on a 5-kg box of coffee resting on a on a horizontal surface. the force of friction on th

e box is 8-N. what is its acceleration and final velocity and its final position if pushed at this rate for 10 seconds
Physics
1 answer:
KatRina [158]3 years ago
3 0

The acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.

Answer:

Explanation:

As per Newton's second law of motion, acceleration of any object is directly proportional to the net external unbalanced force acting on that object and inversely proportional to the mass of the object.

Since there are two forces acting on the box in opposite direction, the net force will be the difference of horizontal and frictional force acting on the object.

Net force = Horizontal force - Frictional force = 18 N - 8 N = 10 N.

Now, from second law of motion, Acceleration = \frac{Net force}{Mass}

So, acceleration = 10 N /5 kg = 2 m/s².

Since, acceleration exerted by the box is found to be 2 m/s², we can determine the final velocity of the object after 10 seconds using the first equation of motion.

v = u + at, Here v is the final velocity and u is the initial velocity which is zero for the present case. Other parameters like a is found to be 2 m/s² and time is 10 seconds.

So Final velocity v = 0+(2×10)=20 m/s.

And the final position can be determined using the second equation of motion.

s = ut+1/2at²

Final position = (0×10)+(0.5×2×10×10)= 100 m.

So the final position is 100 m.

Thus, the acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.

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a bus is moving at 22m/s [E] for 12s. Then the bus driver slows down at 1.2m/s2 [W] until the bus stops. Determine the total dis
KatRina [158]
The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:

d = vt = (22 m/s)(12 s) = 264 m

For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²

Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m

Thus,
Total Displacement = 264 m +  201.67 m = 465.67 or  approximately 4.7×10² m.
7 0
3 years ago
What is the frequency heard by a person driving at 15 m/s toward a factory whistle emitting a
zmey [24]

Answer:

835.29 Hz

Explanation:

When moving towards the source of sound, frequency will be given by

f*=f(vd+v)/v

Where f is the freqiency of the source, vd is the driving speed, v is the speed of sound in air, f* is the inkown frequency when moving forward.

Substituting 800 Hz for f, 340 m/s for v and 15 m/s for vd then

f*=800(15+340)/340=835.29411764704 Hz

Rounded off, the frequency is approximately 835.29 Hz

4 0
3 years ago
Ballistic data obtained on a firing range show that aerodynamic drag reduces the speed of a .44 magnum revolver bullet from 250
m_a_m_a [10]

Answer:

0.363999909622

Explanation:

F = Force

m = Mass = 15.6 g

C = Drag coefficient

ρ = Density of air = 1.21 kg/m³

A = Surface area = \dfrac{\pi}{4}d^2

v = Terminal velocity = v=210\ m/s

s = Displacement = 150 m

a=\dfrac{v^2-u^2}{2s}

Force is given by

F = ma

F=\dfrac{1}{2}\rho CAv^2\\\Rightarrow ma=\dfrac{1}{2}\rho CAv^2\\\Rightarrow m\dfrac{v^2-u^2}{2s}=\dfrac{1}{2}\rho CAv^2\\\Rightarrow C=2\times m\dfrac{v^2-u^2}{2s}\times\dfrac{1}{\rho Av^2}\\\Rightarrow C=2\times15.6\times 10^{-3}\dfrac{210^2-250^2}{2\times 150}\times\dfrac{1}{1.21\times\dfrac{\pi}{4}\times (11.2\times 10^{-3})^2(210)^2}\\\Rightarrow C=-0.363999909622

The drag coefficient is 0.363999909622 (ignoring negative sign)

4 0
3 years ago
Two trains A and B of length 400 m each are moving on two parallel tracks with
Vinvika [58]

<u>Answer:</u>

<em>The initial distance between the trains is 1450 m. </em>

<u>Explanation:</u>

In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.

Train B’s acceleration  is 1m/s^2   and it accelerated for 50 seconds.

<em>a=1 m/s^2</em>

<em>t=50 s </em>

<em>initial speed u=72km/h </em>

<em>we have to convert this speed into m/s  </em>

<em>u=72 \times 5/18=20 m/s</em>

<em>Distance covered in accelerating phase  S=ut+1/2  at^2  </em>

<em>=20 \times 50+1/2 \times 1 \times 50^2</em>

<em>=1000+1250=2250 m </em>

If  a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.

<em>Here length of train A+length of train B=400+400=800 m</em>

<em>Hence the initial distance between the trains = 2250-800=1450 m</em>

6 0
3 years ago
A speed does not involve the element of
guajiro [1.7K]
A speed does not involve the element of direction.
6 0
3 years ago
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