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lubasha [3.4K]
3 years ago
11

An Apple falls from a tree and one-half second later hits the ground

Physics
1 answer:
docker41 [41]3 years ago
8 0
There isnt enough information to answer the question, the missing variable is "distance from said falling spot and ground"
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Which scenario is an application of Newtons Second Law of Motion?
svlad2 [7]
D would be the answer because The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
4 0
3 years ago
The National Aeronautics and Space Administration (NASA) studies the physiological effects of large accelerations on astronauts.
m_a_m_a [10]

Answer:

v = 23.95 m/s

Explanation:

As we know that when astronaut is revolving in circular path then the acceleration of the astronaut is due to centripetal acceleration

so it is given as

a_c = \frac{v^2}{R}

here we know that

a_c = 4.50 g

also we know that

R = 13 m

now we have

4.50 \times 9.81 = \frac{v^2}{13}

v = 23.95 m/s

3 0
4 years ago
Tech A says a camshaft position sensor must be positioned correctly to accurately signal the position of the camshaft to the pow
qaws [65]

Answer:

Both Tech A and Tech B

Explanation:

It's important to position sensor accurately to the position of camshaft to the powertrain control in the ignition module as said by Tech A. Equally, to achieve what Tech A says, a special tool can be used for comparing the position of camshaft sensor to crankshaft sensor. Therefore, both technicians are correct.

6 0
3 years ago
Find the y-component of this vector:73.3,12.0m. Remember, angles are measured from the +x axis.
Rainbow [258]

Answer:

11.5 meters

Explanation:

73.3°, 12.0m

The y-component is the magnitude of the vector times the sine of the angle measured from the +x axis.

y = 12.0 sin (73.3°)

y = 11.5

The y-component is 11.5 meters.

7 0
3 years ago
A spring with k = 19.5 N/cm is initially stretched 1.39 cm from its equilibrium length. a) How much more energy is needed to fur
lara31 [8.8K]

Answer:

Energy needed = 54.02 J

Explanation:

the Energy in an elastic spring from hookes law is given  as

F= ke , therefore the energy (E) is

E = \frac{1}{2} Ke^{2}

K = 19.5 N/cm

e = 1.39cm

E = \frac{1}{2} x 19.5 x 1.39

E = 13.55 J

The energy to stretch the spring for 6.93cm is

E = \frac{1}{2} x 19.5 x 6.93

E = 67.57 J

The more energy needed for the further stretch is

67.57 - 13.55

Energy needed = 54.02 J

7 0
3 years ago
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