All categories

13 days ago

Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+βF(x)=αx3+β.

Physics

1 answer:

Sav [1.1K]13 days ago

8 0

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To calculate the kinetic energy variation, we can utilize the work-energy theorem.

W = ΔK

∫ F .dx = K - K₀

If the object starts from rest, then K₀ = 0.

So, ∫ F dx cos θ = K.

As the force and displacement directions align, the angle is zero, and hence the cosine is 1.

Now we can substitute and perform integration:

α ∫ x³ dx + β ∫ dx = K.

Thus, α x⁴ / 4 + β x = K.

Next, we evaluate from the limits F = 0 to F:

α (x⁴ / 4 - 0) + β (x - 0) = K.

Consequently, K = αX⁴ / 4 + β x.

This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.

To finalize the computation, we need to ascertain the displacement.

You might be interested in

A motorcycle traveling at 36 m/s slams on the brakes to avoid an accident. The motorcycle skids 23m before stoping. What is the

ValentinkaMS [1149]

Since the motorcycle was at a speed of 36 m/s prior to braking, that marks the initial velocity.
$u = 36 {ms}^{ - 1}$
The motorcycle skidded 23m before coming to a full stop, indicating that
$s = 23m$
As it has ceased motion, the final velocity is zero.

$v = 0{ms}^{ - 1}$
We can apply the 'suvat' formula relevant to linear motion.

${v}^{2} = {u}^{2} + 2as$
Substituting the aforementioned values allows us to find,

${0}^{2} = {36}^{2} + 2a(23)$

$0 = 1296+ 46a$
$46a = - 1296$
$a = - 28.2 {ms}^{ - 2}$

We can apply the formula
$v = u + at$
to calculate the time required for the motorcycle to stop.

$0 = 36 + - 28.2t$
$- 36 = - 28.2t$
$t = 1.3s$

8 0

3 days ago

What is the mass of a baseball clocked moving at a speed of 105 mph or 46.9 m/s and wavelength 9.74 × 10-35m?

Maru [1056]

To address this issue, we apply the de Broglie equation written as:

λ = h/mv
where h equals 6.626×10⁻³⁴ J·s

Solving for m, we substitute for v, which is 46.9 m/s:
9.74 × 10⁻³⁵ m = 6.626×10⁻³⁴ J·s / (m)(46.9 m/s)
Thus, we find that m = 0.145kg

6 0

1 day ago

A force of 500 N is exerted on a baseball by the bat for 0.001 s. What is the change in momentum of the baseball?

ValentinkaMS [1149]

Answer: Δp = F*Δt = 500N*0.001s = 0.5Ns

3 0

1 day ago

While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylind

Yuliya22 [1153]

Answer:

Part A: 7.75 m/s

Part B: 2330.8 kN

Part C: 24.03 kN

Part D: 4.8 kN

Part E: $1.7\times 10^{9} Dyn/cm^{2}$

Part F: Option D

Bending one's legs lengthens the duration of force application from the ground, resulting in a reduction of the applied force.

Explanation:

Part A

Using the fundamental kinematic equations

$v^{2}=u^{2}+2gh$ where v represents the velocity just before ground impact, g denotes gravitational acceleration, u signifies initial velocity, and h is the fall height.

With the initial velocity at zero, thus:

$v^{2}=2gh$

$v=\sqrt 2gh$

Plugging in 10 m/s² for g and 3 m for h gives:

$v=\sqrt 2\times 10\times 3 =\sqrt 60= 7.745967\approx 7.75 m/s$

Part B

The force exercised by the leg can be expressed as

F = PA where P is pressure, F indicates force, and A denotes the cross-sectional area of the bone.

$A=\frac {\pi d^{2}}{4}$

With a substitution of 2.3 cm or 0.023m for d and $1.7\times10^{8} N/m2$ for P, we derive the force as:

$F=PA=1.7\times10^{8}*\frac {\pi (0.023)^{2}}{4}= 2330818.276\approx 2330.8 kN$

Part C

The fundamental kinematic equations from part (a) can also be rearranged to show:

$v^{2}=u^{2}+2a\triangle x$ and solving for a yields

$a=\frac {v^{2}-u^{2}}{2\triangle x}$ where a is the acceleration and $\triangle x$ signifies the change in length.

Using the previously derived value from part a, 7.75 m/s for v, and 0.01 m for $\triangle x$ gives us:

$a=\frac {7.75^{2}-0^{2}}{2\times 0.01}= 3003.125 m/s^{2}$

The force felt by the man is given by:

$F=ma=80\times 3003.125= 240250 N\approx 24.03 kN$

Part D

A similar approach with the fundamental kinematic equations shows:

$v^{2}=u^{2}+2a\triangle h$ and solving for a indicates:

$a=\frac {v^{2}-u^{2}}{2\triangle h}$ where a is the acceleration and $\triangle h$ denotes the change in height.

The force experienced can thus be defined as $F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}$ .

For substitution, we use m = 80 Kg, and 0.5m for \triangle h along with other values calculated in part c.

$F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}=80\times \frac {7.75^{2}-0^{2}}{2\times 0.5}= 4805 N\approx 4.8 kN$

Part E

$P=1.7\times 10^{8}=1.7\times 10^{8}\times (\frac {10^{5} Dyn}{10^{4} cm^{2}}=1.7\times 10^{9} Dyn/cm^{2}$

Part F

Bending one's legs extends the period over which the force acts, thus lessening the overall force exerted by the ground.

7 0

4 days ago

A water park is designing a new water slide that finishes with the rider flying horizontally off the bottom of the slide. The sl

serg [1198]

Response: 12.62 meters

Clarification:

3 0

6 days ago