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Zigmanuir [339]
1 year ago
7

Ways in which a teacher plays a role in the literacy development of the learners​

Physics
1 answer:
RideAnS [48]1 year ago
7 0

Answer:

encourage all attempts at reading, writing, and speaking

Explanation:

You might be interested in
An object that is 3 times higher than another object of the same mass will
uranmaximum [27]

Answer: I feel that 3 is the answer

Explanation: Let there be 2 objects, A and B

A is at height of 5m whereas B is at height of 15m

so over here let the gravitational potential energy of A be x

and since B is 3 times higher than A B=3x

Since, earth is considered to be the point where gravitational potenial is 0

So hence forth and object 3 times up will have 3 times the gravitational potential energy of A

3 0
3 years ago
A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west
ivolga24 [154]

Answer:

Magnitude of avg velocity, |v_{avg}| = 18.9 km/h

\theta' = 56.85^{\circ}

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = \frac{27}{60} h

Angle, \theta = 50^{\circ}

Time taken in 50^{\circ}east of due North direction, t' = 29 min =  \frac{29}{60} h

Time taken in west direction, t'' = 37 min =  \frac{27}{60} h

Solution:

Now, the displacement, 's' in east direction is given by:

\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km

Displacement in  50^{\circ} east of due North for 29.0 min is given by:

\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}

\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}

\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km

Now, displacement in the west direction for 37 min:

\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km

Now, the overall displacement,

\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}

\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}

\vec{s_{net}} =  16.03\hat{i} + 24.54\hat{j} km

(a) Now, average velocity, v_{avg} is given:

v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}

v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}

v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h

Magnitude of avg velocity is given by:

|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h

(b) angle can be calculated as:

tan\theta' = \frac{15.83}{10.34}

\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}

6 0
3 years ago
.
insens350 [35]

Answer:

-\frac{1}{2}

Diverging lens

Explanation:

Given parameters:

Power of lens  = -2.0D

Unknown:

Focal length  = ?

Solution:

The power of lens is the reciprocal of the focal length;

   P  = \frac{1}{f}

where  f is the focal length

 f  = \frac{1}{P}   = -\frac{1}{2}

The lens is a diverging lens

8 0
2 years ago
What speed should a satellite of mass 4,900 kg moving around
Rudiy27

Based on the calculations, the speed required for this satellite to stay in orbit is equal to 1.8 × 10³ m/s.

<u>Given the following data:</u>

  • Gravitational constant = 6.67 × 10⁻¹¹ m/kg²
  • Mass of Moon = 7.36 × 10²² kg
  • Distance, r = 4.2 × 10⁶ m.

<h3>How to determine the speed of this satellite?</h3>

In order to determine the speed of this satellite to stay in orbit, the centripetal force acting on it must be sufficient to change its direction.

This ultimately implies that, the centripetal force must be equal to the gravitational force as shown below:

Fc = Fg

mv²/r = GmM/r²

<u>Where:</u>

  • m is the mass of the satellite.
  • M is mass of the Moon.

Making v the subject of formula, we have;

v = √(GM/r)

Substituting the given parameters into the formula, we have;

v = √(6.67 × 10⁻¹¹ × 7.36 × 10²²/4.2 × 10⁶)

v = √(1,168,838.095)

v = 1,081.13 m/s.

Speed, v = 1.8 × 10³ m/s.

Read more on speed here: brainly.com/question/20162935

#SPJ1

8 0
2 years ago
Which refers to the amount of heat required to change the temperature of 1 gram of a substance by 1°C and is related to the chem
timama [110]

Answer:

Specific heat

Explanation:

8 0
2 years ago
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