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2 years ago

7

A car whose total mass is 800kg travelling with a uniform velocity of 20m/s suddenly observes a stationary dog 50m ahead on its

path. If the total force applied on the braking system on the car by the driver is 2000N, what would happen?

1 answer:

maxonik [38]2 years ago

7 0

Answer:

The driver hits the stationery dog because the applied force is less than required force

Explanation:

Kinetic energy will be given by

$KE=0.5mv^{2}$ where m is the mass of the vehicle and v is the speed/velocity of the vehicle.

Substituting 800 Kg for m and 20 m/s for v we obtain

$KE=0.5*800*(20 m/s)^{2}=160,000$

Frictional force by vehicle pads is given by

$Fr=\frac {KE}{d}$ where d is the distance moved

Substituting 160000 for KE and 50 m for d we obtain

$Fr=\frac {160000}{50}=3200 N$

Therefore, the vehicle hits the dog since the required force is 3200N but the driver applied only 2000 N

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The answer will be 50N.
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Which statement provides a complete scientific discription of an object in motion?

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Question: which statement provides a complete scientific discription of an object in motion?

Answer: the marble moved 30 cm north in 6 seconds.

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3 years ago

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The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:

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Answer:

$\frac{d_{1}}{d_{2}}=0.36$

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

$\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]}$ (1)

So, the temperature of the first and the second star will be:

$T_{1}=3866.7 K$

$T_{2}=6444.4 K$

Now the relation between the absolute luminosity and apparent brightness is given:

$L=l\cdot 4\pi r^{2}$ (2)

Where:

L is the absolute luminosity
l is the apparent brightness
r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

$\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}$

So the relative distance between both stars will be:

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}}$ (3)

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σ is the Boltzmann constant
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T is the temperature
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Let's put (4) in (3) for each star.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}$

As we know both stars have the same size we can canceled out the areas.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=0.36$

I hope it helps!

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3 years ago

Two cylinders A&B at the same temperature contains the same quantity of the same kind of gas. Cylinder A has three times the

GREYUIT [131]

Answer:

pressure in cylinder A must be one third of pressure in cylinder B

Explanation:

We are told that the temperature and quantity of the gases in the 2 cylinders are same.

Thus, number of moles and temperature will be the same for both cylinders.

To this effect we will use the formula for ideal gas equation which is;

PV = nRT

Where;

P is prrssure

V is volume

n is number of moles

T is temperature

R is gas constant

We are told that Cylinder A has three times the volume of cylinder .

Thus;

V_a = 3V_b

For cylinder A;

Pressure = P_a

Volume = 3V_b

Number of moles = n

Thus;

P_a × 3V_b = nRT

For cylinder B;

Pressure = P_b

Volume = V_b

Number of moles = n

Thus,

P_b × V_b = nRT

Combining the equations for both cylinders, we have;

P_a × 3V_b = P_b × V_b

V_b will cancel out to give;

3P_a = P_b

Divide both sides by 3 to get;

P_a = ⅓P_b

Thus, pressure in cylinder A must be one third of pressure in cylinder B

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2 years ago

Consider a simple tension member that carries an axial load of P=22.44N. Find the total elongation in the member due to the load

rodikova [14]

Answer:

The total elongation for the tension member is of 0.25mm

Explanation:

Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:

$\sigma=E*\epsilon$ (1)

Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:

$\delta L=L*\epsilon$ (2)

Here L is the member extension and δL the change total longitudinal elongation.

Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:

$\sigma=P/A$

$\sigma=22.44N / 1290 mm^2$

$\sigma=0.0174 N/mm^2$

Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:

$=\sigma=E*\epsilon$

$\epsilon=\sigma/E$

$\epsilon=0.0174 \frac{N}{mm^2}/\ 204 \frac{N}{mm^2}$

$\epsilon=8.53*10^-{5}$

Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):

$\delta L=3048 mm * 8.53*10^{-5}$

$\delta L= 0.25 mm$

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3 years ago