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Nataly [62]
3 years ago
12

What are the five practices in scientific inquiry

Physics
1 answer:
nata0808 [166]3 years ago
5 0
1. Learners are engaged by scientifically oriented questions.
2. Learners give priority to evidence, allowing them to develop and
evaluate explanations that address scientifically-oriented questions.
3. earners formulate explanations form evidence to address scientifically
oriented question
4. Learners evaluate their explanations in light of alternative explanations,
particularly those reflecting scientific understanding.
5. Learners communicate and justify their proposed explanations.
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Beasy is driving in his car to take care of some errands. The first errand has him
lutik1710 [3]

At this point in the story, Beasy has driven his car (2+6+4) = 12 km.

He is parked at the thrift store, (2+4) = 6 km East and 6 km North of his starting point.

As the crow flies, the thrift store is √(6km² + 6km²) in a straight line from the starting point.

That's √(72 km²) , which works out to 8.485 km .  When rounded to the nearest whole km, he can phone up his wife and tell her he's "eight kilometers from home can you hear me now ?".

Displacement is a vector, so to answer the question completely, we also need to state its direction.

The angle from home to the thrift store, relative to East, is arctan(6km/6km).

That's 45 degrees.  

The full displacement vector is <em>8.485 km Northeast.</em>

3 0
3 years ago
The Atwood’s machine shown consists of two blocks of mass m1 and m2 that are connected by a light string that passes over a pull
Talja [164]

(A) For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

(D) For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.

<h3 /><h3>The given parameters:</h3>
  • Mass of block 1 = m1
  • Mass of block 2, = m2
  • Height of block 1 above the ground, = h1
  • Height of block 2 above the ground = h2

The total initial mechanical energy of the two block system is calculated as follows;

m_1gh_1 + \frac{1}{2} m_1v_1_i^2 = m_2gh_2 + \frac{1}{2} m_2v_2_i^2\\\\m_1gh_1 + 0 = m_2gh_2 + 0\\\\m_1gh_1 = m_2gh_2\\\\m_1gh_1 - m_2gh_2 = 0

When the block m2 reaches the ground the block m1 attains maximum height and the total mechanical energy at this point is given as;

m_1g(h_1 + h_2) + K.E_1 = \frac{1}{2}m_2v_{max}^2 + P.E_2\\\\m_1g(h_1 + h_2 ) -PE_2 = \frac{1}{2}m_2v_{max}^2 - K.E_1\\\\m_1g(h_1 + h_2 )  - 0= \frac{1}{2}m_2v_{max}^2 - 0\\\\m_1g(h_1 + h_2 )  = \frac{1}{2}m_2v_{max}^2\\\\W = \frac{1}{2}m_2v_{max}^2

Thus, we can conclude the following before the block m2 reaches the ground;

  • For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
  • For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

5 0
3 years ago
If the tire reaches a temperature of 48 ∘c, what fraction of the original air must be removed if the original pressure of 250 kp
andrey2020 [161]
Assuming air as ideal gas and amount of air in no of moles is known then by gas law,

PV= nRT

Pressure is constant

P* (change in volume) = nR* (change in temperature)
7 0
4 years ago
A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of
SCORPION-xisa [38]

Answer:

a).β=0.53x10^{-3} T

a).β=0.40 x10^{-4} T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

\beta =\frac{u_{o}*I*N}{2\pi*r}

a).

N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A}  \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T

b).

The distance is the radius add the cross section so:

r_{1}=15cm+5cm\\r_{1}=20cm

r_{1} =20cm*\frac{1m}{100cm}=0.20m

\beta =\frac{u_{o}*I*N}{2\pi*r1}

\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T

3 0
3 years ago
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
3 years ago
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