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kari74 [83]
2 years ago
7

A van of mass 5,000kg travelling at a velocity of 10 m/s collides with a stationary car. The vehicles move together after the im

pact at a velocity of 6 m/s. Calculate the mass of the car.
Physics
1 answer:
Lunna [17]2 years ago
5 0

Hi there!

We can use the conservation of momentum for an INELASTIC collision:

m_1v_1 + m_2v_2 = v_f(m_1 + m_2)

Let:

m1 = mass of van (5000 kg)

m2 = mass of car (? kg)

v1 = initial velocity of van (10 m/s)

v2 = initial velocity of car (0 m/s)

vf = final velocity of BOTH objects (6 m/s)

We can rearrange and plug in values to solve for m2:

m_1v_1 + m_2(0) = v_f(m_1 + m_2)\\\\m_1v_1 = v_f(m_1 + m_2)\\\\(5000)(10) = 6(5000 + m_2)

Solve:

50000 = 30000 + 6m_2\\\\20000 = 6m_2\\\\m_2 = 20000/6 = \boxed{3333.33 kg}

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A volumen constante un gas ejerce una presión de 880 mmHg a 20o Celsius dentro de una olla a presión ¿Qué temperatura habrá si e
jasenka [17]

Answer: Hence, the final temperature is 350 K

Explanation :

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=880mmHg\\T_1=20^0C=(20+273)K=293K\\P_2=1050mmHg\\T_2=?

Putting values in above equation, we get:

\frac{880mmHg}{293K}=\frac{1050mmHg}{T_2}\\\\T_2=350K

Hence, the final temperature is 350 K

8 0
2 years ago
Freezing Point Depression: Can someone explain this formula to me? ΔTf = Kfcm
Leya [2.2K]
If the solution is treated as an ideal solution, the extent of freezing point depression depends only on the solute concentration that can be estimated by a simple linear relationship with the cryoscopic constant: ΔTF = KF · m · i ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF (solution). KF, the cryoscopic constant, which is dependent on the properties of the solvent, not the solute. Note: When conducting experiments, a higher KF value makes it easier to observe larger drops in the freezing point. For water, KF = 1.853 K·kg/mol.[1] m is the molality (mol solute per kg of solvent) i is the van 't Hoff factor (number of solute particles per mol, e.g. i = 2 for NaCl).
8 0
3 years ago
In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron
slega [8]

Answer:

a)   v_average = 11 m / s, b)  t = 0.0627 s

, c)    F = 7.37 10⁵ N

, d)   F / W = 35.8

Explanation:

a) truck speed can be found with kinematics

         v² = v₀² - 2 a x

The fine speed zeroes them

           a = v₀² / 2x

           a = 22²/2 0.69

           a = 350.72 m / s²

The average speed is

           v_average = (v + v₀) / 2

           v_average = (22 + 0) / 2

           v_average = 11 m / s

b) The average time

          v = v₀ - a t

          t = v₀ / a

          t = 22 / 350.72

          t = 0.0627 s

c) The force can be found with Newton's second law

             F = m a

             F = 2100 350.72

             F = 7.37 10⁵ N

.d) the ratio of this force to weight

             F / W = 7.37 10⁵ / (2100 9.8)

             F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

5 0
2 years ago
Which goes into which category?
Pavel [41]

Evidence of Chemical Change:

A car rusting

Leaves changing in October

The glow of a light bulb

A white cloudy substance occurs after mixing two substances

Burning toasts

A candle being lit

Evidence of Physical Change:

Water boiling

Yellow and blue paints mixed together

Wax melting

Crushing a rock with a hammer

Hope this helps!

Have a great day!

6 0
3 years ago
When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the
Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

\theta = 54.6^{\circ}

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = \sqrt{\frac{gx}{sin2\theta}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s

Now,

The angular velocity can be calculated as:

v = \omega R

Thus

\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s

8 0
3 years ago
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