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3 years ago

a car moving east at 45 km/h turns and travels west at 30 km/h. What is the magnitude and direction of the change in velocity?

Physics

2 answers:

slega [8]3 years ago

7 0

Answer:

The magnitude and direction of the change in velocity is 75 m/s in west.

Explanation:

Given that,

Velocity of car = 45 km/h in east

Velocity of car = 30 km/h in west

We need to calculate the magnitude of the change in velocity

Using formula of change in velocity

Change in velocity = final velocity - initial velocity

$\Delta v=V_{f}-v_{i}$

Put the value into the formula

$\Delta v=-30-45$

$|\Delta v|=75\ m/s$

The direction of the the change in velocity is toward west.

Hence, The magnitude and direction of the change in velocity is 75 m/s in west.

PSYCHO15rus [73]3 years ago

3 0

Magnitude of change is 15km/h
change in direction is to the right?

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Answer:

100 ÷ 9.58 = 10.44 (approximate answer)

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3 years ago

To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl

Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁ + 2.43 x 0 = v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂² = 0.07032

v₂² = 0.07032 / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

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3 years ago

What is the wavelength of a wave with a frequency of 466 Hz and a speed of

Volgvan

Answer:

<h3>The answer is option B</h3>

Explanation:

The wavelength of a wave can be found by using the formula

$\lambda = \frac{c}{f} \\$

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$\lambda = \frac{343}{466} \\ = 0.73605150...$

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<h2>Answer:</h2>

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