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sukhopar [10]
2 years ago
9

if m represents mass in kg, v represents speed in m/s, and r represents radius in m, show that the force F in the equation F=mv^

2 /r can be expressed in the unit kg×m/s^2
Physics
1 answer:
Zarrin [17]2 years ago
4 0
This approach is called the dimensional analysis which involves only the units of measurement without their magnitudes. You simply have to do the operations by using variables. Cancel out like items that may appear both in the numerator and denominator side. The solution is as follows:

F = mv²/r = [kg][m/s]²/[m] = [kg][m²⁻¹][1/s²] = [kg·m/s²]
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Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68
den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, y_0=1500\ m

Initial horizontal velocity is, u_x=688\ m/s

Initial vertical velocity is, u_y=0(\textrm{Horizontal velocity only initially)}

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is S=y-y_0=0-1500=-1500\ m

Acceleration in the vertical direction is due to gravity, g=-9.8\ m/s^2

Therefore, the displacement of the bomb is given as:

S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s

So, the bomb will remain in air for 17.5 s before hitting the ground.

6 0
2 years ago
An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its
FromTheMoon [43]

Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

F_{drag}=12v+4v^2

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

F_{drag}=mg

12v+4v^2=80\times 9.8

4v^2+12v=784

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.

6 0
2 years ago
would an elephant standing on one leg exert a higher force on a scale than an elephant on four legs. why​
zlopas [31]

Answer:

no becaus force is mass multiplied by acceleration. the mass of the elephant does not change

7 0
2 years ago
A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards
Mrrafil [7]

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

5 0
2 years ago
An 1876 N crate is being pushed across a level force at a constant speed by a force of 747 N. What is the coefficient of kinetic
nekit [7.7K]

The crate only moves horizontally, so its net vertical force is 0. The only forces acting in the vertical direction are the crate's weight (pointing downward) and the normal force of the surface on the crate (pointing upward). By Newton's second law, we have

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0   →   <em>n</em> = <em>mg</em> = 1876 N

where <em>n</em> is the magnitude of the normal force.

In the horizontal direction, the crate is moving at a constant speed and thus with no acceleration, so it's completely in equilibrium and the net horizontal force is also 0. The only forces acting on it in this direction are the 747 N push (pointing in the direction of the crate's motion) and the kinetic friction opposing it (pointing in the opposite direction). By Newton's second law,

∑ <em>F</em> (horizontal) = 747 N - <em>f</em> = 0   →   <em>f</em> = 747 N

The frictional force is proportional to the normal force by a factor of the coefficient of kinetic friction, <em>µ</em>, such that

<em>f</em> = <em>µn</em>   →   <em>µ</em> = <em>f</em> / <em>n</em> = (747 N) / (1876 N) ≈ 0.398188 ≈ 0.40

8 0
2 years ago
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