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a trcuk weighs four times as much as a stationary car. if teh truck coasts into the car at 12 km/s and they stick toegther, what is their final velocity

Answers

Answer 1

Answer:

Answer:

v=9.6 km/s

Explanation:

Given that

The mass of the car = m

The mass of the truck = 4 m

The velocity of the truck ,u= 12 km/s

The final velocity when they stick = v

If there is no any external force on the system then the total linear momentum of the system will be conserve.

Pi = Pf

m x 0 + 4 m x 12 = (m + 4 m) x v

0 + 48 m = 5 m v

5 v = 48

$v=(48)/(5)\ km/s$

v=9.6 km/s

Therefore the final velocity will be 9.6 km/s.

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You have devised an experiment to measure the kinetic coefficient of friction between a ramp and block. You place the block on the ramp at an angle high enough that it starts sliding. You measure the time it takes to fall down a known distance. The time it takes to fall down the ramp starting from a standstill is 0.5 sec, ???? = 1 kg, θ = 45o, and the distance it falls, L, is 0.5 m. What is µk? (8 pts)

Answers

Answer:

μ = 0.423

Explanation:

To solve this exercise we must use Newton's second law and kinematics together, let's start using expressions of kinematics to find the acceleration of the body

Let's fix a reference system where the x axis is parallel to the inclined plane, but the acceleration is only on this axis

x = v₀ t + ½ a t²

The body starts from rest so its initial speed is zero

a = 2 x / t²

a = 2 0.5 /0.5²

a = 4 m / s²

Taking the acceleration of the body, we use Newton's second law, we take the direction up the plane as positive

X axis

fr - Wₓ = m a (1)

Y Axis

N- $W_(y)$ = 0

N = W_{y}

We use trigonometry to find the components of the weight

sin 45 = Wₓ / W

cos 45 = W_{y} / W

Wₓ = W sin 45

W_{y} = W cos 45

The out of touch has the expression

fr = μ N

fr = μ W_{y}

We substitute in 1

μ mg cos 45 - mg sin 45 = m a

W_{y} = (a + g sin 45) / g cos 45

μ = a / g cos 45 + 1

We calculate

Acceleration goes down the plane, so it is negative

a = -4 m / s²

μ = 1- 4 / (9.8 cos 45)

μ = 0.423

Answer:

The μ = 0.422

Explanation:

The distance travelled by the mass is equal to:

$L=ut+(1)/(2)at^(2) \n0.5=(0*5)+(1)/(2) a(0.5^(2) )\na=4m/s^(2)$

The sum of forces in y-direction equals zero:

∑Fy = 0

N - (m * g * cosθ) = 0

N - (1 * 9.8 * cos45) = 0

N = 6.93 N

The sum of forces in x-direction is equal to:

∑Fx = ma

(m * g * sinθ) - fk = m * a

(1 * 9.8 * sin45) - fk = 1 * 4

fk = 2.93 N

fk = μ * N

2.93 = μ * 6.93

μ = 0.422

Vaporized gases and released dust form a bright cloud called a(n) ____ around the solid part of a comet.

Answers

The mixture of gases round the body or nucleus of the comet is called a coma.

the right answer is gonna be Coma

Part A A microphone is located on the line connecting two speakers hat are 0 932 m apart and oscillating in phase. The microphone is 2 83 m from the midpoint of the two speakers What are the lowest two trequencies that produce an interflerence maximum at the microphone's location? Enter your answers numerically separated by a comma

Answers

Answer:

a) $f=368.025\ \textup{Hz}$

b) $f_2=736.051\ \textup{Hz}$

Explanation:

Given:

The distance between two speakers (d) = 0.932 m

The distance of the microphone from the midpoint = 2.83 m

Thus, distance of microphone from the nearest speaker (L) = 2.83 - (0.932/2) = 2.364 m

also, the distance of the microphone from the farther speaker (L') = 2.83 + (0.932/2) = 3.296 m

Now,

The path difference is calculated as

L' - L = d = 0.932 m

Now,for a maxima to be produced at the microphone, the waves must constructively interfere.

for this to happen the path difference should be integral multiple of the wavelength.

thus,

$\textup d = n\lambda$

hence, the largest wavelength will be for n = 1,

therefore,

0.932 = 1 × λ

λ = 0.932 m

now, the velocity of sound is given as c = 343 m/s

thus, the frequency will be

$f=(c)/(\lambda)$

on substituting the values, we get

$f=(343)/(0.932)=368.025\ \textup{Hz}$

now, the 2nd largest wavelength will be for n = 2

0.932 = 2 × λ

λ = 0.466

thus, the frequency will be

$f_2=(343)/(0.466)=736.051\ \textup{Hz}$

hence, these are the lowest first two frequencies.

A block-spring system consists of a spring with constant k = 475 N/m attached to a 2.50 kg block on a frictionless surface. The block is pulled 5.50 cm from equilibrium and released from rest. For the resulting oscillation, find the amplitude, angular frequency, frequency, and period. What is the maximum value of the block's velocity and acceleration?

Answers

Explanation:

It is given that,

Spring constant of the spring, k = 475 N/m

Mass of the block, m = 2.5 kg

Elongation in the spring from equilibrium, x = 5.5 cm

(a) We know that the maximum elongation in the spring is called its amplitude. So, the amplitude for the resulting oscillation is 5.5 cm.

(b) Let $\omega$ is the angular frequency. It is given by :

$\omega=\sqrt{(k)/(m)}$

$\omega=\sqrt{(475)/(2.5)}$

$\omega=13.78\ rad/s$

(c) Let T is the period. It is given by :

$T=(2\pi)/(\omega)$

$T=(2\pi)/(13.78)$

T = 0.45 s

(d) Frequency,

$f=(1)/(T)$

$f=(1)/(0.45)$

f = 2.23 Hz

(e) Let v is the maximum value of the block's velocity. It is given by :

$v_(max)=\omega* A$

$v_(max)=13.78* 5.5* 10^(-2)$

$v_(max)=0.757\ m/s$

The value of acceleration is given by :

$a=\omega^2A$

$a=(13.78)^2* 5.5* 10^(-2)$

$a=10.44\ m/s^2$

Hence, this is the required solution.

Final answer:

The amplitude of the block-spring system is 0.055 m, with an angular frequency of 13.77 rad/s, and a frequency of approximately 2.19 Hz. The system has a period of approximately 0.46 s, with the maximum velocity being 0.76 m/s, and the maximum acceleration being 189 m/s².

Explanation:

In this block-spring system, we can determine the oscillation properties with the given parameters: mass (m = 2.50 kg), spring constant (k = 475 N/m), and displacement (x = 5.50 cm = 0.055 m).

Amplitude (A): This is the maximum displacement from the equilibrium position, in this case, it is 0.055 m, the distance from which the block is pulled and released.
Angular Frequency (ω): This can be calculated by the formula ω = √(k/m) which is equal to √(475/2.50) = 13.77 rad/s.
Frequency (f): It is given by the formula f = ω / 2π ≈ 2.19 Hz.
Period (T): The time for one complete cycle of the motion, calculated as T = 1/f ≈ 0.46 s.
Maximum value of velocity (v max): Calculated by the formula v = ω*A = 13.77 * 0.055 = 0.76 m/s.
Maximum value of acceleration (a max): Maximum acceleration occurs when the block is at the maximum displacement (amplitude), found with formula a = ω²*A = 189 m/s².

Learn more about Oscillations here:

brainly.com/question/30111348

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When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.300 V. What is the internal resistance of this battery

Answers

The internal resistance of the battery is 0.5 ohms.

To calculate the internal resistance of the battery, we use the formula below

Formula:

(V/R)r = V'............. Equation 1

Where:

V = Voltage across the terminal of the battery
R = Resistance connected across the battery
r = internal resistance of the battery
V' = voltage drop of the battery.

Make r the subject of the equation

r = V'R/V............ Equation 2

From the question,

Given:

V = 12 V
R = 20 ohms
V' = 0.3 V

Substitute these values into equation 2

r = (0.3×20)/12
r = 6/12
r = 0.5 ohms.

Hence, The internal resistance of the battery is 0.5 ohms.

Learn more about internal resistance here: brainly.com/question/14883923

Answer:

The internal resistance is $r = 0.5 \ \Omega$

Explanation:

From the question we are told that the resistance of

The resistance of the resistor is $R = 20.0\ \Omega$

The voltage is $V = 12.0 \ V$

The magnitude of the voltage fall is $e = 0.300\ V$

Generally the current flowing through the terminal due to the voltage of the battery is mathematically represented as

$I = (V)/(R)$

substituting values

$I = (12.0 )/(20 )$

$I = 0.6 \ A$

The internal resistance of the battery is mathematically represented as

$r = (e)/(I)$

substituting values

$r = (0.300)/( 0.6 )$

$r = 0.5 \ \Omega$

what is the necessary condition on a force the result the conservation of angular momentum for a particle affected by that force?

Answers

Answer:

The force must be applied on the axis of rotation

Explanation:

A rotating system conserves its angular momentum only if there are no external torques on the system. In other words, the external torque must be equal to zero.

T=0

T=Fxd

Torque is equal to the vector product of a force by the distance between the axis of rotation and where the force is applied.

For this product to be zero, the force must be applied on the axis of rotation (d=0).

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