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3 years ago

If a 2,000-kg car hits a tree with 500 n of force over a time of 0.5 seconds, what is the magnitude of its impulse?

Physics

2 answers:

timurjin [86]3 years ago

8 0

Answer:

The magnitude of its impulse is 250 N-sec.

Explanation:

Given that,

Mass of car = 2000 kg

Force = 500 N

Time = 0.5 sec

We need to calculate the impulse

Using formula of impulse

$I = F\times\delta t$

Where, F = force

$\Delta t$ = time

Put the value into the formula

$I=500\times0.5$

$I=250\ N-s$

Hence, The magnitude of its impulse is 250 N-sec.

Elden [556K]3 years ago

4 0

Impulse = Force * time
Impulse = 500N *0.5 s =250 N*s

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A metal conduit will be used as a pathway for wiring through a concrete block. The conduit is a 4 foot long rod with an outer di

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Answer: $510. 12 cm^3$

The volume of a cylindrical conduit, $V=\pi (r_2^2-r_1^2) h$

where $r_1$ is the inner radius, $r_2$ is the outer radius, h is the length.

First of all, we would covert each parameter into centimeters.

length of the conduit, $h=4 ft=4\times 30.48 cm=121.92 cm$

outer radius, $r_1=2.2/2 in =1.1 in= 1.1\times 2.54 cm=2.79 cm$

inner radius, $r_1=2/2 in =1 in= 2.54 cm$

Insert the values in the volume formula,

$V=3.14 \times (2.79^2-2.54^2) \times 121.92 cm^3=510. 12 cm^3$

Hence, the volume of metal in the conduit is 510. 12 cubic centimeters.

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3 years ago

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Why is the gravitational potential energy of an object 1 meter above the moon’s surface less than its potential energy 1 meter a

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Potential energy relative to the surface is

(mass) x (acceleration due to gravity) x (height above the surface).

At 1.0 meter above the surface, that is

 (mass) x (acceleration due to gravity) x (1.0 meter) .

The object's mass doesn't change, so the only thing that has any effect
on its potential energy at 1 meter above the surface is the acceleration
of gravity or, in other words, the surface of what ?

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3 years ago

A 16 g rifle bullet traveling 240 m/s buries itself in a 3.6 kg pendulum hanging on a 2.5 m long string, which makes the pendulu

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Answer:

x = 0.54 m

y = 0.058 m

Explanation:

m = mass of the bullet = 16 g = 0.016 kg

v = speed of bullet before collision = 240 m/s

M = mass of the pendulum = 3.6 kg

L = length of the string = 2.5 m

h = height gained by the pendulum after collision

V = speed of the bullet and pendulum combination

Using conservation of momentum

m v = (m + M) V

(0.016) (240) = (0.016 + 3.6) V

V = 1.062 m/s

Using conservation of energy

Potential energy gained by bullet and pendulum combination = Kinetic energy of bullet and pendulum combination

(m + M) g h = (0.5) (m + M) V²

(9.8) h = (0.5) (1.062)²

h = 0.058 m

y = vertical displacement = h = 0.058 m

x = horizontal displacement

horizontal displacement is given as

x = sqrt(L² - (L - h)²)

x = sqrt(2.5² - (2.5 - 0.058)²)

x = 0.54 m

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3 years ago

Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar

brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

At State 1:

The first step is to find the vapor pressure

$P_{v1} = \rho_1 P_g_1$

$= \phi_1 P_{x \ at \ 125^0C}$

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

$P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1$

$116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)$

$116 \times 10^3 \times v_{v_1} = 183831.7778$

$v_{v_1} = 1.584 \ m^3/kg$

At State 1:

The next step is to determine the mass of water vapor pressure.

$m_{v1} = \dfrac{V}{v_{v1}}$

$= \dfrac{2.5}{1.584}$

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air $m_a$ $P_{a1} V = m_a \dfrac{\overline R}{M_a}T_1$

$(P_1-P_{v1}) V = m_a \dfrac{\overline R}{M_a}T_1$

$(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)$

$710000= m_a \times 114220.642$

$m_a = \dfrac{710000}{114220.642}$

$m_a = 6.216 \ kg$

For the specific volume $v_{v_1} = 1.584 \ m^3/kg$ , we get the identical value of saturation temperature

$T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)$

$T_{sat} =101.92 ^0\ C$

Thus, at $T_{sat} =101.92 ^0\ C$ , condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

At State 1;

The internal energy is calculated as:

$U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )$

At $T_1$ = 125° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 125 ^0C } = 278.93 + ( 286.16 -278.93) (\dfrac{398-390}{400-390} )$

$=278.93 + ( 7.23) (\dfrac{8}{10} )$

$= 284.714 \ kJ/kg\\$

At $T_1$ = 125° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 125^0C} = u_g = 2534.5 \ kJ/kg$

$U_1 = (m_a u_a \ at \ _{ 125 ^0C }) + ( m_{v1} u_v \ at \ _{125^0C} )$

= 6.216 × 284.714 + 1.578 × 2534.5

= 5768.716 kJ

At State 2:

The internal energy is calculated as:

$U_2 = (m_a u_a \ _{ at \ 110^0 C})+ ( m_{v1} u_v \ _{ at \ 110^0 C} )$

At temperature 110° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 110^0C } = 271.69+ ( 278.93-271.69) (\dfrac{383-380}{390-380} )$

$271.69+ (7.24) (0.3)$

$= 273.862 \ kJ/kg\\$

At temperature 110° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 110^0C}= 2517.9 \ kJ/kg$

$U_2 = (m_a u_a \ at \ _{ 110 ^0C }) + ( m_{v1} u_v \ at \ _{110^0C} )$

= 6.216 × 273.862 + 1.578 × 2517.9

= 5675.57 kJ

Finally, the heat transfer during the process is

Q = U₂ - U₁

Q = (5675.57 - 5768.716 ) kJ

Q = -93.146 kJ

with the negative sign, this indicates that heat is lost from the system.

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3 years ago

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Yes that is correct. We know this because 4.00 x 10 4 Pa is constant. If you have 2.00×10−3m3 then you do the following: (2.00×10^−3)(4.00×10^ 4) = 8.00×10^−3. That is how you get your answer

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3 years ago

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