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

750 miles / hour

Explanation:

velocity = distance / time

= 4500 miles / 6 hours

= 750 miles / hour

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.)

## Related Questions

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I can't really tell what (I) is, but F) blender def converts electric into motion of spinning blades

Which describes one feature of the image formed by a convex mirror?????

The image formed by a convex mirror will always have its smaller than the size of the object no matter what the position of the object.

Explanation:

The image formed by a convex mirror will always have its smaller than the size of the object no matter what the position of the object.

Also notice that convex mirror always makes virtual images.

Another feature of the convex mirror is that an upright image is always formed by the convex mirror.

An important mirror formula to remember which is applicable for both convex and mirrors

• 1/f= 1/u + 1/v

Here:

'u' is an object which gets placed in front of a spherical mirror of focal

length 'f' and image 'u' is formed by the mirror.

right side up

Explanation:

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.250 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.960 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor. 4.33 Correct: Your answer is correct. m/s (b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.

a. 3.73 m/s b. 27.8 m/s²

Explanation:

(a) Calculate his velocity (in m/s) when he leaves the floor.

Using the conservation of energy principles,

Potential energy gained by basketball player = kinetic energy loss of basket ball player

So, ΔU + ΔK = 0

ΔU = -ΔK

mg(h' - h) = -1/2m(v'² - v²)

g(h' - h) = -1/2(v'² - v²) where g = acceleration due to gravity = 9.8 m/s², h' = 0.960 m, h = 0.250 m, v' =0 m/s (since the basketball player momentarily stops at h' = 0.960 m) and v = velocity with which the basketball player leaves the floor

Substituting the values of the variables into the equation, we have

9.8 m/s²(0.960 m - 0.250 m) = -1/2((0 m/s)² - v²)

9.8 m/s²(0.710 m) = -1/2(-v²)

6.958 m²/s² = v²/2

v² = 2 × 6.958 m²/s²

v² = 13.916 m²/s²

v = √(13.916 m²/s²)

v = 3.73 m/s

(b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.

Using v² = u² + 2as where u = initial speed of basketball player before lengthening = 0 m/s, v = final speed of basketball player after lengthening = 3.73 m/s, a = acceleration during lengthening and s = distance moved during lengthening = 0.250 m

So, making, a subject of the formula, we have

a = (v² - u²)/2s

Substituting the values of the variables into the equation, we have

a = ((3.73 m/s)² - (0 m/s)²)/(2 × 0.250 m)

a = (13.913 m²/s² - 0 m²/s²)/(0.50 m)

a = 13.913 m²/s²/(0.50 m)

a = 27.83 m/s²

a ≅ 27.8 m/s²

At what distance from a long straight wire carrying acurrentof 5.0A is the magnitude of the magnetic field due to the
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?

The distance is 2 cm

Solution:

According to the question:

Magnetic field of Earth, B_{E} =

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

where

d = distance from current carrying wire

Now,

d = 0.02 m 2 cm

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.)

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.

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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.

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The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. 1.11 × 104 Pa 1.09 × 105 Pa 1.13 × 107 Pa 1.11 × 108 Pa 2.18 × 105 Pa