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Mariana [72]
2 years ago
7

How can goal setting help with academic performance?

Physics
1 answer:
san4es73 [151]2 years ago
6 0

Answer:

Setting goals keeps students focused on desired outcomes and provides a clear direction for success

You might be interested in
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t
marishachu [46]

Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete (\Delta t), in seconds:

\Delta t = t_{A}-t_{C} (1)

Where:

t_{C} - Time spent by the sound in concrete, in seconds.

t_{A} - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}

\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}}  \right)

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  } (2)

Where:

v_{C} - Speed of the sound in concrete, in meters per second.

v_{A} - Speed of the sound in the air, in meters per second.

x - Distance traveled by the sound, in meters.

If we know that \Delta t = 6.4\,s, v_{C} = 3000\,\frac{m}{s} and v_{A} = 343\,\frac{m}{s}, then the distance travelled by the sound is:

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  }

x = 2478.585\,m

The impact occured at a distance of 2478.585 meters from the person.

7 0
2 years ago
What is the velocity of a proton that has been accelerated by a potential difference of 15 kV? (i)\:\:\:\:\:9.5\:\times\:10^5\:m
andre [41]

Answer:

Velocity of a proton, v=1.7\times 10^6\ m/s    

Explanation:

It is given that,

Potential difference, V=15\ kV=15\times 10^3\ V

Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.

Using the conservation of energy as :

\dfrac{1}{2}mv^2=qV

q is the charge of proton

m is the mass of proton

v=\sqrt{\dfrac{2qV}{m}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}

v=1695361.75\ m/s

v=1.69\times 10^6\ m/s

or

v=1.7\times 10^6\ m/s

So, the velocity of a proton is 1.7\times 10^6\ m/s. Hence, this is the required solution.

8 0
3 years ago
a spiral spring of natural length 10.0cm has a scale pan hanging freely in its tower ends . When an object of mass of 20g is pla
weqwewe [10]

Answer:

Explanation:

There seems to be a typo in the problem statement.  It says the spring stretches to a shorter length after more mass is added.  Please check the problem statement.  I'm going to do the calculations assuming that the first length should be 11.80 cm and the second length should be 12.05 cm.

Hooke's law states that the force needed to compress or extend a linear spring is:

F = kΔx, where k is the stiffness and Δx is the displacement.

When a 20g object is placed in the pan, the spring stretches to a length of 11.80 cm.  The force of the spring is counteracting the weight of both the pan and the object. Therefore:

(m + 0.020) g = k (0.1180 - 0.100)

And when another 30g object is placed in the pan, the spring stretches to a length of 12.05 cm.

(m + 0.020 + 0.030) g = k (0.1205 - 0.100)

We now have two equations and two variables.  If we divide the second equation by the first equation:

(m + 0.050) / (m + 0.020) = (0.1205 - 0.100) / (0.1180 - 0.100)

(m + 0.050) / (m + 0.020) = 0.02050 / 0.0180

0.0180 (m + 0.050) = 0.02050 (m + 0.020)

0.0180 m + 0.0009 = 0.02050 m + 0.00041

0.00049 = 0.0025 m

m = 0.196

The pan has a mass of 0.196 kg, or 196 g.

4 0
2 years ago
Given the nuclear equation:
Leokris [45]
This is a nuclear fission reaction, in which a larger nucleus is bombarded with a neutron to make it break down into two smaller nuclei and release energy.
6 0
2 years ago
The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn
insens350 [35]

Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

A_x=Acos\theta       (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N

the y comonent of the vecor is Ay = 49.92 N

3 0
2 years ago
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