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Ratling [72]
3 years ago
5

A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s

outh. Imagine that both of these vehicles crash into each other head-on and get stuck together. What will their velocity be immediately after the collision
Physics
1 answer:
trasher [3.6K]3 years ago
8 0

Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum p_{i} must be equal to the final momentum p_{f}, and taking into account this is aninelastic collision:

Before the collision:

p_{i}=mV_{o}+MU_{o} (1)

After the collision:

p_{f}=(m+M)V_{f} (2)

Where:

m=1380 kg is the mass of the car

V_{o}=23 m/s is the velocity of the car, directed to the north

M=1625 kg is the mass of the truck

U_{o}=-26 m/s is the velocity of the truck, directed to the south

V_{f} is the final velocity of both the car and the truck

p_{i}=p_{f} (3)

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg} (6)

Finally:

V_{f}=-3.49 m/s The negative sign indicates the direction of the velocity is to the south

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Two large aluminum plates are separated by a distance of 2.0 cm and are held at a potential difference of 170 V. An electron ent
notsponge [240]

Answer:

Explanation:

Electric field between plates

V / d

= 170 / ( 2 x 10⁻² )

= 8500 N/C

Force on electron in this field

= 8500 x 1.6 x 10⁻¹⁹

= 13600 x 10⁻¹⁹ N

Acceleration

= 13600 x 10⁻¹⁹ / 9.1 x 10⁻³¹

a = 1494.5 x 10¹² m /s²

s = .1 x 10⁻² m

v² = u² + 2as

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= 8.41 x 10¹⁰ + 299 x 10¹⁰

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8 0
3 years ago
A 400-n block is dragged along a horizontal surface by an applied force as shown. the coefficient of kinetic friction is uk = 0.
gulaghasi [49]
The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.

We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force F_f acting against the motion. Since their resultant must be zero, we have:
F-F_f = 0
The frictional force is
F_f = \mu mg
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mg=400 N is the weight of the block. 

Substituting these values, we find the magnitude of the force F:
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4 0
3 years ago
1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position
antiseptic1488 [7]

Answer:

The value of the distance is \bf{14.52~cm}.

Explanation:

The velocity of a particle(v) executing SHM is

v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)

where, \omega is the angular frequency, A is the amplitude of the oscillation and x is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., x = 0.

The maximum velocity(\bf{v_{m}}) is

v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Divide equation (1) by equation(2).

\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)

Given, v = 0.25 v_{m} and A = 15~cm. Substitute these values in equation (3).

&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm

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Explain more. I can't understand

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You rapidly swirl a mixture of substances with different densities
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4 years ago
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