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Lana71 [14]
3 years ago
5

Sound waves require a ________ to travel through

Physics
2 answers:
Karolina [17]3 years ago
7 0

Answer:

sound Requires a medium in order to travel.

Nimfa-mama [501]3 years ago
6 0

Answer:

medium

Explanation:

this is because sound waves need particles to transfer energy from one point to another

hope this helps

please mark it

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A child is pushing a merry-go-round. The angle through which the merry-go-round has turned varies with time according to θ(t)=γt
Lana71 [14]

Answer:

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Explanation:

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7 0
3 years ago
explain the production of current with the help of a magnet and describe which device works on this phenomena
spin [16.1K]

Answer:

Explanation:

The properties of magnets are used to make electricity. Moving magnetic fields pull and push electrons. ... Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current.

4 0
2 years ago
A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is
Alex17521 [72]

Hello!

We can use the following equation for calculating power dissipated by a resistor:
P = i^2R

P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)

Plug in the known values and solve.

P = (2.0^2)(50) = \boxed{\text{ B. }200 W}

7 0
2 years ago
Vector A⃗ points in the negative y direction and has a magnitude of 5 km. Vector B⃗ has a magnitude of 15 km and points in the p
Alexxandr [17]

Answer:

magnitude of A − B =  15.81 km

Explanation:

Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a  magnitude of 15 km.

According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.

A(0,-5)

B(15,0)

A - B = (-15 i - 5 j )

Magnitude of the vector is given by

|A - B| = \sqrt{(-15)^{2}+(-5)^{2}}

|A - B| = \sqrt{250}

|A - B| = 15.81 km

7 0
3 years ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
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