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3 years ago

With what speed does the can move immediately after the collision? Answer in units of m/s.

Physics

1 answer:

Ratling [72]3 years ago

7 0

Answer:

1.74 m/s

Explanation:

From the question, we are given that the mass of the an object, m1= 2.7 kilogram(kg) and the mass of the can,m(can) is 0.72 Kilogram (kg). The velocity of the mass of an object(m1) , V1 is 1.1 metre per seconds(m/s) and the velocity of the mass of can[m(can)], V(can) is unknown- this is what we are to find.

Therefore, using the formula below, we can calculate the speed of the can, V(can);

===> Mass of object,m1 × velocity of object, V1 = mass of the can[m(can)] × velocity is of the can[V(can)].----------------------------------------------------(1).

Since the question says the collision was elastic, we use the formula below

Slotting in the given values into the equation (1) above, we have;

1/2×M1×V^2(initial velocity of the first object) + 1/2 ×M(can)×V^2(final velocy of the first object)= 1/2 × M1 × V^2 m( initial velocity of the first object).

Therefore, final velocity of the can= 2M1V1/M1+M2.

==> 2×2.7×1.1/ 2.7 + 0.72.

The velocity of the can after collision = 1.74 m/s

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3 years ago

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The following information are given in the question:
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3 years ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

2 years ago

Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien

otez555 [7]

The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s

The height at which the block stops rising is approximately 1.1415 m

The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

$Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m} = \mathbf{ 1000\, N/m}$

Coefficient of kinetic friction, $\mu_k$ = 0.39

Therefore, we have;

Friction force = $\mathbf{\mu_k}$ ·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, <em>W</em> ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

$\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d$

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, <em>W</em> ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

The velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>

The height at which the block will stop moving, <em>h</em>, is given as follows;

$At \ the \ maximum \ height, \ h, \ we \ have ; \ \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x$

Which gives;

$Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{ 7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2 } \approx 1.536$

The distance up the inclined, the block rises, at maximum height is therefore;

$x_{max}$ ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

The height at which the block stops rising, h ≈ <u>1.1415 m</u>

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

Length of the incline, $x_{max}$ = 1.536 m

Learn more about conservation of energy here:

brainly.com/question/7538238

5 0

2 years ago