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What is the speed of an airplane that travels 4500 miles in 6 hours?

Answers

Answer 1

Answer:

Answer:

750 miles / hour

Explanation:

velocity = distance / time

= 4500 miles / 6 hours

= 750 miles / hour

Answer 2

Answer:

Given:-

Speed of the airplane (s) = 4500miles
Time taken (t) = 6h

ToFind:Speed (v) of the particle (airplane).

We know,

s=vt

where,

s = Distance,
v = Speed &
t = Time taken.

Similarly,

v=s/t

→ v = (4500 miles)/(6 hours)

→ v = 750 miles/hour ...(Ans.)

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Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. The resultant vector A+ B is given by

Answers

Answer:

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

Explanation:

Let $\vec A = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})$ and $\vec B = 4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})$ , both measured in meters. The resultant vector $\vec R$ is calculated by sum of components. That is:

$\vec R = \vec A+\vec B$ (Eq. 1)

$\vec R = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})+4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})$

$\vec R = (6\cdot \cos 30^(\circ)-4\cdot \sin 30^(\circ))\,\hat{i}+(6\cdot \sin 30^(\circ)-4\cdot \cos 30^(\circ))\,\hat{j}$

$\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?

Answers

Answer:

$v=3.564\ m.s^(-1)$

$\Delta v =2.16\ m.s^(-1)$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2* (9.8* (1.8)/(3))* 3$

$v_J=5.94\ m.s^(-1)$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30* 5.94+0=(30+20)v$

$v=3.564\ m.s^(-1)$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30* 9.8* (1.8)/(3)$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=(F_R)/(m_J)$

$a_j=(71.4)/(30)$

$a_j=2.38\ m.s^(-2)$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2* 2.38* 3$

$v_J_n=3.78\ m.s^(-1)$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^(-1)$

What If? What would be the new angular momentum of the system (in kg · m2/s) if each of the masses were instead a solid sphere 15.0 cm in diameter? (Round your answer to at least two decimal places.)

Answers

Final answer:

To find the new angular momentum of the system if each of the masses were solid spheres, calculate the moment of inertia for each sphere using the formula (2/5) × m × r^2. Multiply the moment of inertia of each sphere by the angular velocity of the system to find the new angular momentum.

Explanation:

The angular momentum of a system can be found by multiplying the moment of inertia of the system with its angular velocity.

If each of the masses were instead a solid sphere 15.0 cm in diameter, we would need to calculate the moment of inertia of each sphere using the formula for the moment of inertia of a solid sphere, I = (2/5) × m × r^2, where m is the mass and r is the radius of the sphere.

Once we have the moment of inertia for each sphere, we can multiply it by the angular velocity of the system to find the new angular momentum.

Learn more about Angular momentum here:

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Final answer:

The new angular momentum, given the same angular speed, will be 0.9 times the original, as the moment of inertia for the system is replaced with that of solid spheres of given mass and radius.

Explanation:

The question is asking for the new angular momentum of a sphere with a given diameter if we replace each of the masses in a given system with it. To compute the new angular momentum, it's crucial to recognize that angular momentum (L) is given by the product of the moment of inertia (I) and angular velocity (w). The moment of inertia for a solid sphere is given by (2/5)mr^2, where m is the mass and r is the radius of the sphere. Since angular velocity has not been specified in the question, it would be assumed to remain unchanged.

So, for this specific system, each mass is replaced with a solid sphere of mass 20 kg and radius 15 cm (or 0.15 m). Thus using the formula for solid sphere inertia, I = (2/5)*(20 kg)*(0.15 m)^2 = 0.9 kg*m^2. If w remains the same, then the new angular momentum L = I * w will be 0.9 times the original angular momentum. This is because w is the same but the moment of inertia has a new value due to the shape and size of the new masses.

Learn more about Angular Momentum here:

brainly.com/question/37906446

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Based on the measured force between objects that are 10 meters apart, how can you find the force between objects that are any distance apart ?

Answers

The force between objects that are any distance apart is expressed as $P'=(100P)/(r^2)$

According to the gravitational law, the force acting on an object is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically,

$P=(GMm)/(r^2)$

M and m are the masses

r is the distance between the masses

If the force between objects that are 10 meters apart, hence;

$P=(GMm)/(10^2)\nP=(GMm)/(100)\nGMm = 100P$

To find the force between objects that are any distance apart, we will use the same formula above to have;

$P'=(GMm)/(r^2)\n$

Substitute the result above into the expression to have:

$P'=(100P)/(r^2)$

Hence the force between objects that are any distance apart is expressed as $P'=(100P)/(r^2)$

Learn more on gravitational law here: brainly.com/question/11760568

Answer:

F' = 100 F/r²

Explanation:

The gravitational force of attraction between two objects is given by the Newton's Gravitational Formula. The Newton's Gravitational Formula is as follows:

F = Gm₁m₂/r²

where,

F = Force between objects

G = Universal Gravitational Constant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects = 10 m

Therefore,

F = Gm₁m₂/10²

Gm₁m₂ = 100F --------------------- equation (1)

Now, we consider these objects at any distance r apart. So, the force becomes:

F' = Gm₁m₂/r²

using equation (1), we get:

F' = 100 F/r²

So, if the force (F) between objects 10 m apart is known, we can find it at any distance from the above formula.

At what partial pressure are argon atoms expected to have a free travel of approximately 5 µm, if the gas is at a temperature of 400 K? The cross section of collision, σ, or Argon is 0.28 nm2Ar molar mass is 39.9 g/mole

Answers

Answer:

2790 Pa

Explanation:

Given wavelength λ= 5μm

temperature T= 400 K

cross section of collision σ= 0.28 nm^2

molar mass = 39.9 g/mole

pressure = $P= (RT)/(√(2)N_A\sigma\lambda )$

putting values we get

= $(8.314*400)/(√(2)*6.022*10^(23)*0.28*10^(-18)*5*10^(-6) )$

⇒P = 2790 J/m^3

the partial pressure are argon atoms expected= 2790 Pa

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate from 2.00 to 6.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.