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3 years ago

5

The collision between a hammer and a nail can be considered to be approximately elastic. Estimate the kinetic energy acquired by

a 13 g nail when it is struck by a 550 g hammer moving with an initial speed of 4.6 m/s.

1 answer:

AlekseyPX3 years ago

3 0

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Energy cannot be created nor destroyed. A. Ella said "a lot of energy in the hot water has disappeared." Explain why her stateme

harkovskaia [24]

Answer:

See Explanation

Explanation:

The principle of conservation of energy states that; energy can neither be created nor destroyed but is converted from one form to another.

In view of this principle, Ella can not be correct when she says that a lot of energy has disappeared. The use of the term "disappeared" connotes the idea that the energy no longer exists which does not happen.

Hence, energy can not "disappear" from hot water rather the energy in the water may be transferred to the surroundings.

6 0

2 years ago

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

2 years ago

A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 ms from an initial spe

dezoksy [38]

Answer:

the magnitude of the average contact force exerted on the leg is 3466.98 N

Explanation:

Given the data in the question;

Initial velocity of hand v₀ = 5.25 m/s

final velocity of hand v = 0 m/s

time interval t = 2.65 ms = 0.00265 s

mass of hand m = 1.75 kg

We calculate force on the hand F $_{hand$

using equation for impulse in momentum

F $_{hand$ × t = m( v - v₀ )

we substitute

F $_{hand$ × 0.00265 = 1.75( 0 - 5.25 )

F $_{hand$ × 0.00265 = 1.75( - 5.25 )

F $_{hand$ × 0.00265 = -9.1875

F $_{hand$ = -9.1875 / 0.00265

F $_{hand$ = -3466.98 N

Next we determine force on the leg F $_{leg$

Using Newton's third law of motion

for every action, there is an equal opposite reaction;

so, F $_{leg$ = - F $_{hand$

we substitute

F $_{leg$ = - ( -3466.98 N )

F $_{leg$ = 3466.98 N

Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N

5 0

2 years ago

How does the toaster serve as an example of Convection energy transfer?

blsea [12.9K]

Check bing for the answer

3 0

3 years ago

A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th

Marysya12 [62]

Answer:

Mininmum acceleration required = $2.74 m/s^{2}$

Explanation:

Given,

Lift off speed = 125km/hr.

We know km/hr conversion factor to m/s is $\frac{5}{18}$ ,

Therefore,

Lift off speed required = $125 \times\frac{5}{18}$ m/s

Lift off speed required = 34.72222 m/s

Length of Takeoff Run given = 220 m.

So,

the flight needs to achieve its takeoff speed at a constant acceleration before it reaches the end of the 220 m runway.

Consider the formula from kinematics,

$v^{2}-u^{2}=2as$ ,

where,

v - final velocity of particle,

u - initial velocity of particle

a - the constant acceleration at which the particle is moving with

s - distance travelled by particle

So at minimum acceleration,it reaches the takeoff speed at the end of the runway.

Therefore in the formula,

we use, v = Lift off speed = 34.72222 m/s;

u = 0 m/s (plane starts from rest);

a = minimum acceleration of the plane

s = 220 m;

Substituting these values in the formula, we get,

$(34.72222)^{2}-0 = 2\times a \times 220$

a = $\frac{(34.72222)^{2} }{2\times220}$

a = $2.74 m/s^{2}$ .

Therefore,the minimum acceleration of the plane required = $2.74 m/s^{2}$ .

7 0

2 years ago