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mixer [17]
2 years ago
10

A car that is traveling in a straight line at 40 km/h can brake to a stop within 20 m. If the same car is traveling at 120 km/h

what would be its stopping distance in this case? Assume the braking force is the same in both cases and ignore air resistance.
Physics
1 answer:
stiks02 [169]2 years ago
8 0

Answer:

180 m

Explanation:

Case 1.

U = 40 km/h = 11.1 m/s, V = 0, s = 20 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0 = 11.1 × 11.1 - 2 × a × 20

a = 3.08 m/s^2

Case 2.

U = 220 km/h = 33.3 m/s, V = 0

a = 3.08 m/s^2

Let the stopping distance be x.

Again use third equation of motion

0 = 33.3 × 33.3 - 2 × 3.08 × x

X = 180 m

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Gas and oil from the dead creatures, the gas, and oil are trapped under layers of rock
scoundrel [369]

Answer:

Gas is on top of oil

Explanation:

Gas is less dense ( smaller density ) than oil which is more dense. Since density is directly proportional to mass and inversely proportional to volume, gas will occupy a larger space than oil.

The larger space occupied is the space above the mixture of oil and gas, hence gas is above oil

8 0
3 years ago
A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnit
gayaneshka [121]

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m(v-u).................. Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

I = 2.4(4)

I = 9.6 Ns

4 0
3 years ago
An airplane has wings, each with area A, designed so that air flows over the top of the wing at 265 m/s and underneath the wing
exis [7]

Answer

given,

Pressure on the top wing = 265 m/s

speed of underneath wings = 234 m/s

mass of the airplane =  7.2 × 10³ kg

density of air =  1.29 kg/m³

using Bernoulli's equation

 P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2

 \Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)

 \Delta P =\dfrac{1}{2}\times 1.29\times (265^2-234^2)

 \Delta P =9977.5 Pa

Applying newtons second law

2 Δ P x A - mg = 0

A =\dfrac{mg}{2\Delta P}

A =\dfrac{7.2\times 10^3 \times 9.8}{2\times 9977.5}

    A = 3.53 m²

7 0
3 years ago
An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elect
beks73 [17]

Answer:

Speed of the alpha particle is v=1.8180\times 10^3m/sec      

Explanation:

We have given charge on alpha particle q=3.2\times 10^{-19}C

Mass of the alpha particle m=6.68\times 10^{-27}kg

Potential difference V=-3.45\times 10^{-3}volt

We have to find the speed of the alpha particle

From energy conservation we know that

\frac{1}{2}mv^2=qV

\frac{1}{2}\times 6.68\times 10^{-27}\times v^2=3.2\times 10^{-19}\times 3.45\times 10^{-3}

v=1.8180\times 10^3m/sec

4 0
3 years ago
A small solid sphere and a small thin hoop are rolling along a horizontal surface with the same translational speed when they en
torisob [31]
This situation has a basis such that the solid sphere and the hoop has the same mass. The analysis could be made<span> backwards . The ball will decelerate fastest, so not roll as high. The sphere will accelerate faster, but this also means it decelerates faster on the way up. Hence the answer is the hoop if the masses are equal </span>
3 0
3 years ago
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