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Zigmanuir [339]
2 years ago
7

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

Physics
1 answer:
RideAnS [48]2 years ago
7 0

Answer:

encourage all attempts at reading, writing, and speaking

Explanation:

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The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m
BARSIC [14]

A) The resultant force is 30.4 N at 25.3^{\circ}

B) The resultant force is 18.7 N at 43.9^{\circ}

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

F_1 = 20 N at 0^{\circ} above x-axis

F_2 = 15 N at 60^{\circ} above y-axis

Resolving each force:

F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N

F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N

So, the components of the resultant are:

F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N

And the magnitude of the resultant is:

F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N

And the direction is:

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

\theta=180^{\circ}-60^{\circ}=120^{\circ}

So we have:

F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N

So, the components of the resultant this time are:

F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N

And the magnitude is:

F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N

And the direction is:

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0
3 years ago
Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of
Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m  

we know,

c = f × λ  

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency  

thus,

f=\frac{c}{\lambda}

substituting the values in the equation we get,

f=\frac{3.0\times 10^8 m/s}{2.9m}

f = 1.03 x 10⁸Hz  

Now,

The time period (T) = \frac{1}{f}

or

T =  \frac{1}{1.03\times 10^8}  = 9.6 x 10⁻⁹ seconds  

thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s  

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s  

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}  

Thus, the minimum distance to target = \frac{290}{2} = 145 m

6 0
3 years ago
Why does the speed of sound depend on air temperature?
IRISSAK [1]
<span>The speed of sound is dependent on how close together the molecules of the transmitting medium is.</span>
4 0
3 years ago
Read 2 more answers
Which element on the Periodic Table has fewer than 8 electrons but still has a full valence shell?
lora16 [44]
The answer to this is Helium :) it's in the farthest right columb and is a noble gas.

please mark as brainliest!
6 0
3 years ago
Read 2 more answers
A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The d
blsea [12.9K]

The average power needed to produce this final speed is 1069.1 hp.

Mass of the dragster,  m = 500.0 kg,

Displacement travelled by the dragster,  s = 402 m,

Time taken in this travel,  t = 5.0 s,

Final velocity of the dragster,  v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation,  s = ut + (1/2)at^2.

402  =  u*5  + (1/2)*a*5^2

10*u + 25*a  = 804.      ...........(1)

Using kinematical equation, v = u +at.

130 = u + 5*a

5*u + 25*a = 650.       .............(2)

Solving (1)and (2), we get,

u =  30.8 m/s.

According to work-energy theorem,

Work done = change in kinetic energy

W  = (1/2)*m*(v^2 - u^2)

W = (1/2)*500*(130^2 - 30.8^2)

W  =  3987840. J

Therefore power rating of the dragster is given by,

P  ⇒  W/t. =  3987840/5 = 797568 watt.

P  ⇒ 797568/746 =  1069.1 hp.

Learn more about Power rating here brainly.com/question/20137708

#SPJ4

5 0
2 years ago
Read 2 more answers
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