1, 2, & 3.........................

If force remains constant and acceleration decreases what must happen to the mass

blondinia [14]

That can only be happening if the mass mysteriously increased somehow. I'd like to know how in the world THAT happened.

4 0

3 years ago

A force F→=(cx-3.00x2)iˆ acts on a particle as the particle moves along an x axis, with F→ in newtons, x in meters, and c a cons

DerKrebs [107]

Answer:

Explanation:

Work done = ∫Fdx

= ∫(cx-3.00x²) dx

[ c x² / 2 - 3 x³ / 3 ]₀²

= change in kinetic energy

= 11-20

= - 9 J

[ c x² / 2 - x³ ]₀² = - 9

c x 2² / 2 - 2³ = -9

2c - 8 = -9

2c = -1

c = - 1/2

6 0

4 years ago

Read 2 more answers

A daredevil is shot out of a cannon at 31.4 ◦ to the horizontal with an initial speed of 25.9 m/s. A net is positioned a horizon

Tanzania [10]

Answer:

The net should be placed at a height of 8.45 m

Explanation:

In order to calculate the height you have to apply the equations of <em>projectil motion.</em>

For x-axis, the position at a time t is given by:

Xf=Xo +(VoCosα)t (I)

Where Xf is the final position, Xo is the initial position, Vo is the initial speed and t is the time.

For the y-axis, the position at a time t is given by:

Yf=Yo + (VoSinα)t - 0.5g $t^{2}$ (II)

Where Yf is the final position, Yo is the initial position, Vo is the initial speed, t is the time and g is the acceleration of gravity.

For both equations, ∝ is the angle formed by the cannon and the horizontal

The height where the net should be placed in order to catch the daredevil is given by replacing the time when the daredevil will be in the horizontal position of 39.7 m in the equation of position for the y-axis.

So, you have to calculate the time using (I)

Let the system of reference be placed at x=o and y=o

Xf= 0 + 25.9Cos(31,4◦)t

Xf=39.7

39.7=22.1t

Dividing both sides by 22.1 to find the value of t:

t=1.79 seconds

Replacing the time in equation (II) to find the height:

Yf=0 +25.9Sin(31,4◦)(1.79) - 0.5(9.8) $1.79^{2}$

Yf= 8.45 m

So the height is 8.45 m

4 0

3 years ago

You are given two balls that are made from the same rubber. They are also the same size and color. One is hollow and one is soli

Sveta_85 [38]

Weight, firmness, bounce.

Those would be my best guesses.

3 0

4 years ago

Read 2 more answers

The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alt

lesantik [10]

Answer:

(a) $\triangle l=5 mm$

(b) 0.033

Explanation:

(a)

Force F=mg where m is mass and g is acceleration due to gravity whose value is taken as $9.81 m/s^{2}$

However, for this case, the maximum force is 8 times the weight of runner hence F=8mg

Assume Young's modulus for tendon is $0.15*10^{10} N/m^{2}$

Young's modulus is given by

$E=\frac {Fl}{A\triangle l}$ and $\triangle l=\frac {Fl}{EA}$ and substituting F with 8mg we obtain $\triangle l=\frac {8mgl}{EA}$

Where E is young's modulus, l is stretched length and \triangle l is change in length

Substituting m as 70 kg, g as $9.81 m/s^{2}$ , l as 15cm=0.15 m, E as $0.15*10^{10} N/m^{2}$ and A as $110 m^{2}=0.000110 m^{2}$

$\triangle l=\frac {8*70 Kg*9.81 m/s^{2}*0.15m}{0.15*10^{10} N/m^{2} *0.00011 m^{2}}=0.004994182 m$

$\triangle l=5 mm$

(b)

Strain, $\epsilon=\frac {\triangle l}{l}$ and the fraction of tendon’s length is the ratio of change in length to the stretched length

The fraction of tendon, f is given by

$f=\frac {\triangle l}{l}$ . Substituting $\triangle l$ with 0.005m and l with 0.15m we obtain

$\epsilon=f=\frac {0.005}{0.15}=\frac {1}{30}=0.033$

Therefore, fraction of the tendon’s length is 0.033

5 0

4 years ago