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Tanya [424]
3 years ago
11

A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. If the mass is set in motion in a medium that

imparts a damping force numerically equal to 16 times the velocity, the correct differential equation for the position x (t ), of the mass at a function of time, t is
Physics
1 answer:
denpristay [2]3 years ago
6 0

Answer:

d²x/dt² = - 4dx/dt - 4x is the required differential equation.

Explanation:

Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,

F = mg

kx = mg

k = mg/x

= 4 kg × 9.8 m/s²/2.45 m

= 39.2 kgm/s²/2.45 m

= 16 N/m

Now the drag force f = 16v where v is the velocity of the mass.

We now write an equation of motion for the forces on the mass. So,

F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass

-kx - 16v = 4a

-16x - 16v = 4a

16x + 16v = -4a

4x + 4v = -a where v = dx/dt and a = d²x/dt²

4x + 4dx/dt = -d²x/dt²

d²x/dt² = - 4dx/dt - 4x which is the required differential equation

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A bicyclist starts at rest and speeds up to 30 m/s while accelerating at 4 m/s^2. Determine the distance traveled.
raketka [301]

Answer:

Distance, d = 112.5 meters

Explanation:

Initially, the bicyclist is at rest, u = 0

Final speed of the bicyclist, v = 30 m/s

Acceleration of the bicycle, a=4\ m/s^2

Let s is the distance travelled by the bicyclist. The third equation of motion is given as :

v^2-u^2=2as

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

s=\dfrac{(30)^2}{2\times 4}

s = 112.5 meters

So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.

6 0
2 years ago
A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

3 0
3 years ago
How long a star lives and what it becomes at the end of its life depends primarly on what
liraira [26]
I believe the answer would be mass. Low mass stars and medium mass stars often become white dwarfs when they die while high mass stars explode in violent explosions called supernovas and usually leave behind a black hole or a neutron star.
4 0
3 years ago
The body weighing 8kg moves in a straight line uniformly accelerated with an acceleration of 3m / s? on a horizontal surface, un
Usimov [2.4K]

Answer:

pls explain this to me

Explanation:

I will not answer it

3 0
2 years ago
An automobile engine delivers 47.4 hp. how much time will it take for the engine to do 6.82 × 105 j of work? one horsepower is e
kogti [31]
1 horsepower is equal to 746 W, so the power of the engine is
P=47.4 hp \cdot 746  \frac{W}{hp}=35360 W
The power is also defined as the energy E per unit of time t:
P= \frac{E}{t}
Where the energy corresponds to the work done by the engine, which is E=6.82 \cdot 10^5 J. Re-arranging the formula, we can calculate the time t needed to do this amount of work:
t =  \frac{E}{P}= \frac{6.82 \cdot 10^5 J}{35360 W}=19.3 s
8 0
3 years ago
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