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shtirl [24]
3 years ago
14

HELPP!!

Physics
1 answer:
lisov135 [29]3 years ago
8 0

Answer:

1.2 MPS

Explanation:

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The table below lists the speed that sound travels through four different materials.One of these materials is a gas, two are liq
Firdavs [7]

In order to find the solid, you would want the object in which sound travels the fastest

In this case, since in object C, the speed of sound is the fastest, it is the most likely to be a solid

So object C is most likely to be a solid

5 0
1 year ago
Describe the steps in solving the problem below.
12345 [234]

The velocity of the red cart after the collision is 2 m/s

From the law of conservation of momentum, initial momentum of system = final momentum of system.

m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄ where m₁ = mass of red cart = 4 kg, v₁ = velocity of red cart before collision = + 4 m/s, v₃ = velocity of red cart after collision, m₂ = mass of blue cart = 1 kg, v₂ = velocity of blue cart before collision = 0 m/s (since it is initially at rest) and v₄ = velocity of blue cart after collision = + 8 m/s.

Substituting the values of the variables into the equation, we have,

m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄

4 kg × 4 m/s + 1 kg × 0 m/s = 4v₃ + 1 kg × 8 m/s

16 kgm/s + 0 kgm/s =  4v₃ + 8 kgm/s

16 kgm/s =  4v₃ + 8 kgm/s

16 kgm/s - 8 kgm/s =  (4 kg)v₃

(4 kg)v₃ = 8 kgm/s

Divide both sides by 4 kg, we have

v₃ = 8 kgm/s ÷ 4 kg

v₃ = 2 m/s

The velocity of the red cart after the collision is 2 m/s.

Learn more about conservation of momentum here:

brainly.com/question/7538238

3 0
3 years ago
What is the relationship in which the ratio of the manipulated variable and the responding variable is constant?
son4ous [18]
That's a 'direct' or 'proportional' variation.

The variables are related by the equation      Y = K X .

That constant ratio is the number    ' K ' .

5 0
3 years ago
A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in
Kruka [31]

Answer:

1020g

Explanation:

Volume of can=1100cm^3=1100\times 10^{-6}m^3

1cm^3=10^{-6}m^3

Mass of can=80g=\frac{80}{1000}=0.08kg

1Kg=1000g

Density of lead=11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3

By using 1g/cm^3=10^3kg/m^3

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can  and lead inside it they are floating in the water.

Buoyancy force =F_b=Weight of can+weight of lead

\rho_wV_cg=m_cg+m_lg

Where \rho_w=10^3kg/m^3=Density of water

m_c=Mass of can

m_l=Mass of lead

V_c=Volume of can

Substitute the values then we get

1000\times 1100\times 10^{-6}=0.08+m_l

1.1-0.08=m_l

m_l=1.02 kg=1.02\times 1000=1020g

1 kg=1000g

Hence, 1020 grams of lead shot can it carry without sinking water.

4 0
4 years ago
Two 1 MHz radio antennas emitting in-phase are separated by600
algol [13]

Answer:

1 km

Explanation:

d_0 = Gap between antennas = 600 m

\nu = Frequency = 1 MHz

z_1 = Distance to receiver = 2 km

c = Speed of light = 3\times 10^8\ m/s

Wavelength is given by

\lambda=\dfrac{c}{\nu}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1\times 10^6}\\\Rightarrow \lambda=300\ m

Distance to be moved is given by

D=\dfrac{z_1\lambda_0}{d_0}\\\Rightarrow D=\dfrac{2\times 300}{600}\\\Rightarrow D=1\ km

The distance to be moved is 1 km north.

6 0
3 years ago
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