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3 years ago

What do you think would happen if a lamp close to the sensor was left turned on overnight?

Physics

2 answers:

LenKa [72]3 years ago

8 0

The sensor would go off

kirza4 [7]3 years ago

4 0

They would turn off automatically to save the battery. Hope it helped!

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The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle betwee

lara [203]

Answer:51.44 units

Explanation:

Given

x component of vector is $-27.3\hat{i}$

y component of vector is $43.6\hat{j}$

so position vector is

$r=-27.3\hat{i}+43.6\hat{j}$

Magnitude of vector is

$|r|=\sqrt{27.3^2+43.6^2}$

$|r|=\sqrt{2646.25}$

|r|=51.44 units

Direction

$tan\theta =\frac{43.6}{-27.3}=-1.597$

vector is in 2nd quadrant thus

$180-\theta =57.94$

$\theta =122.06^{\circ}$

4 0

3 years ago

A sports car is advertised to be able to stop in a distance of 50.0 m from a speed of 80 km. What is its acceleration and how ma

Flauer [41]

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car. It can be calculated using third equation of motion as :

$v^2-u^2=2ad$

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{0-(22.22)^2}{2\times 80000}$

$a=-0.00308\ m/s^2$

Value of g, $g=9.8\ m/s^2$

$a=\dfrac{-0.00308}{9.8}\ m/s^2$

$a=(-0.000314)\ g\ m/s^2$

Hence, this is required solution.

8 0

3 years ago

What is transferred by a radio wave?

Alex

Answer: B. energy

Explanation:

4 0

3 years ago

Suppose a uniform electric field of 4 N/C is in the positive x direction. When a charge is placed at and fixed to the origin, th

Yuliya22 [10]

Answer:

E_total = 3 N / A

Explanation:

The electric field is a vector magnitude so when adding we must use vectors, in this case as the initial field E = 4N / c goes towards the axis axis and the field created by the fixed charge (E1) is also on the axis x we can add in scalar form.

E_total = E + E₁

the expression for the field of a point charge is

E₁ = k q₁ / r²

for the point x = 2m, they do not say that the total field is zero, so the charge q1 must be negative

E_total = E -k q₁ / r₂

we substitute

0 = E - k q₁ / r²

q₁ = $\frac{E r^2}{k}$

let's calculate

q₁ = $\frac{4 \ 2^2}{9 \ 10^{-9}}$

q₁ = 1.78 10⁻⁹ C

now we can calculate the field for position x = 4 m

E_total = 4 - 9 10⁹ 1.78 10⁻⁹ / 4²2

E_total = 3 N / A

8 0

3 years ago

A race car has a centripetal acceleration of 13.33 m/s^2 as it goes around a curve. if the curve is a circle with a radius 30 m

anzhelika [568]

Answer:

The speed of the car, v = 19.997 m/s

Explanation:

Given,

The centripetal acceleration of the car, a = 13.33 m/s²

The radius of the curve, r = 30 m

The centripetal force acting on the car is given by the formula

F = mv²/r

Where v²/r is the acceleration component of the force

a = v²/r

Substituting the values in the above equation

13.33 = v²/30

v² = 13.33 x 30

v² = 399.9

v = 19.997 m/s

Hence, the speed of the car, v = 19.997 m/s

3 0

3 years ago