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3 years ago

11

In the parallel spring system, the springs are positioned so that the 37 N weight stretches each spring equally. The spring cons

tant for the left-hand spring is 2.7 N/cm and the spring constant for the righ-hand spring is 4.3 N/cm. How far down will the 37 N weight stretch the springs?

1 answer:

Reptile [31]3 years ago

3 0

Answer:

x = 5.29 m

Explanation:

given,

weight of stretch = 37 N

left-hand spring constant (k₁)= 2.7 N/cm

right hand spring constant(k₂)= 4.3 N/ cm

spring are connected in parallel

F = F₁ + F₂

F = k₁x + k₂x

F = (k₁+ k₂)x

37= (4.3+ 2.7)x

7 x = 37

x = 5.29 m

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3.27 moles of an ideal gas in a 50.0 L tank has a pressure of 171000 Pa. What is the temperature of the gas? (Unit=degrees C)

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The temperature of the gas is 41.3 °C.

Answer:

The temperature of the gas is 41.3 °C.

Explanation:

So on combining the Boyle's and Charles law, we get the ideal law of gas that is PV=nRT. Here P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature. The SI unit of pressure is atm. So we need to convert 1 Pa to 1 atm, that is 1 Pa = 9.86923× $10^{-6}$ atm. Thus, 171000 Pa = 1.6876 atm.

We know that the gas constant R = 0.0821 atmLMol–¹K-¹. Then the volume of the gas is given as 50 L and moles are given as 3.27 moles.

Then substituting all the values in ideal gas equation ,we get

1.6876×50=3.27×0.0821×T

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the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force

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Answer:

PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

$F_{net} = 51 N$

now we also know that friction force due to ground is given as

$F_f = 24 N$

now we have

$F_{net} = F_{ext} - F_f$

$51 = F_{ext} - 24$

$F_{ext} = 75 N$

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

$a = \frac{F_f}{m}$

$a = \frac{24}{24} = 1 m/s^2$

now we can find the distance by kinematics

$v_f^2 - v_i^2 = 2 a d$

$0 - 1.5^2 = 2(-1)d$

$d = 1.125 m$

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A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

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Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

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b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

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$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

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$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

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