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Setler [38]
3 years ago
5

4 uses of plane mirror​

Physics
1 answer:
faust18 [17]3 years ago
5 0

Answer:

they are used in periscopes,for signalling,in kaleidoscopes,to see round dangerous bends

Explanation:

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a hockey puck with a mass of 0.11 kg is at rest on the horizontal frictionless surface of the rink. a player applies a horizonta
stira [4]

The solution to this ques is available in the image.

Given,

Force= 1N

Mass= 0.11kg

Time= 5sec

Force= mass X accelaration

Accelaration= velocity/ time

Speed=distance/ time

Hence, the speed is 45 m/s and the distance is 225 m.

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8 0
1 year ago
If the mass of a material is 42 grams and the volume of the material is 15 cm^3, what would the density of the material be?
Fantom [35]

Density = mass / volume

Density = 42g / 15cm^3

Density = 2.8g/cm^3

3 0
3 years ago
A 44-turn rectangular coil with length ℓ = 17.0 cm and width w = 8.10 cm is in a region with its axis initially aligned to a hor
Mumz [18]

Answer:

The maximum induced emf in the rotating coil  = 29.66V

The induced emf in the rotating coil when (t = 1.00 s) = 26.66V

The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s

Explanation:

Lets state the parameters we are being given right from the question:

Number of rectangular coil, (N) = 44

Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m

Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m

Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T

Angular Speed of Coil, (ω) = 64 rad/s

(a)

To calculate the induced emf in the rotating cell,we can use the formula:

emf = NBAωsin(ωt)

For maximum induced emf, the value of sin(ωt) will be 1

emf_max = NBAω ; if (A = l × w) , we have:

emf_max  = NB(l × w)ω

subsitituting the parameters into the above equation; we have:

emf_max  = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64

= 29.66V

(b)

At t = 1s, the induced emf is calculated as:

emf = NBAωsin(ωt)

substituting the parameters into the equation, we have:

emf =   44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)

=26.66V

(c)

To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.

i.e emf_max = N\frac{d∅}{dt}

since emf_max = 29.66 and N = 44; we have:

29.66 =  44\frac{d∅}{dt}

\frac{d∅}{dt} = \frac{29.66}{44}

= 0.674 Wb/s

5 0
3 years ago
What type of subsurface material is able to store groundwater?
notka56 [123]
<span>Porous material has many spaces that can hold(store) groundwater.</span>
4 0
2 years ago
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
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