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Alex Ar [27]
3 years ago
15

Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December

10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.2 s and was brought jarringly back to rest in only 1 s. Calculate his (a) magnitude of acceleration in his direction of motion and (b) magnitude of acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. g g
Physics
1 answer:
musickatia [10]3 years ago
6 0

Answer:

    a = 5.53 g ,   a = -15g

Explanation:

This is an exercise in kinematics.

a) Let's look for the acceleration

         as part of rest v₀ = 0

          v = v₀ + a t

           a = v / t

           a = 282 / 5.2

          a = 54.23 m / s²

in relation to the acceleration of gravity

          a / g = 54.23 / 9.8

          a = 5.53 g

b) let's look at the acceleration to stop

         va = 0

         0 = v₀ -2 a y

         a = vi / y

         a = 282/2 1

         a = 141 m /s²

         a / G = 141 / 9.8

          a = -15g

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The bob (weight) at the end of a pendulum has a mass of 0.3 kilograms. The bob is pulled to position B and allowed to swing. It
Ivahew [28]

Answer:

0.147 J

Explanation:

The total energy that has been transformed into thermal energy is equal to the loss of gravitational potential energy between the initial situation (bob at h=0.5 m above the ground) and the final situation (bob back but at h=0.45 m above the ground).

Therefore, we have

E_{thermal}=\Delta U=mgh_1 - mgh_2 = mg(h_1 -h_2)

where

m = 0.3 kg is the mass of the bob

g = 9.8 m/s^2

h1 = 0.5 m is the initial height

h2 = 0.45 m is the final height

Substituting, we find the thermal energy

E_{thermal}=(0.3 kg)(9.8 m/s^2)(0.5 m-0.45 m)=0.147 J

Therefore, the energy transformed into thermal energy is 0.147 J.

3 0
3 years ago
Find the velocity at which the bowling ball must move for its kinetic energy to be equal to that of the meteorite.
Vikentia [17]

Answer:

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

Explanation:

BOOM!!!

7 0
3 years ago
Explain how Copernicus concluded that stars were farther away than planets. Draw a diagram showing how this principle applies to
Rainbow [258]
<span>Copernicus decided this with more of an educated guess than anything. For example is when your standing right next to a plane it's huge Right? Well when it's flying it looks really small. He used the same reasoning for stars. Since it looks small it must be farther away.</span>
8 0
3 years ago
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
2 years ago
"what is the period of one vibration of this tone?"
Viktor [21]
The full question is:
On a keyboard, you strike middle C, whose frequency is 256 Hz. What is the period of one vibration of this tone?
The period of a vibration is the time it takes for the particle to make one full oscillation. Frequency is by definition number of full oscillations per unit of time.
When the frequency is expressed in Hz that unit of time is one second.
So there is the following relation between frequency and period:
T=\frac{1}{f}
When we plug in the numbers we get:
T=\frac{1}{256}=0.0039$s
7 0
3 years ago
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