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kati45 [8]
3 years ago
15

An automobile having a mass of 1,000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring with cons

tant 5.75 ✕ 10^6 N/m and is compressed 3.12 cm as the car is brought to rest. What was the speed of the car before impact, assuming no energy is lost in the collision with the wall?
Physics
1 answer:
Nady [450]3 years ago
4 0

To solve this problem it is necessary to apply the concepts related to the conservation of energy, specifically the potential elastic energy against the kinetic energy of the body.

By definition this could be described as

PE = KE

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

Where

k = Spring constant

x = Displacement

m = mass

v = Velocity

This point is basically telling us that all the energy in charge of compressing the spring is transformed into the energy that allows the 'impulse' seen in terms of body speed.

If we rearrange the equation to find v we have

v = \sqrt{\frac{kx^2}{m}}

Our values are given as

m = 1000kg

k = 5.75*10^6N/m

x = 3.12*10^{-2}m

Replacing at our equation we have then,

v = \sqrt{\frac{kx^2}{m}}

v = \sqrt{\frac{(5.75*10^6)(3.12*10^{-2})^2}{1000}}

v = 2.3658m/s

Therefore he speed of the car before impact, assuming no energy is lost in the collision with the wall is 2.37m/s

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geniusboy [140]

Answer:

\tau = 7.63 Nm

Explanation:

As we know that moment of force is given as

\tau = \vec r \times \vec F

now we have

\vec r = 1.2 m

\vec F = 14 N

now from above formula we have

\tau = r F sin\theta

here we know that

\theta = 27 degree

so we have

\tau = (1.2)(14) sin27

\tau = 7.63 Nm

3 0
3 years ago
What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne
irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance h= 1790 km = 1.790\times10^{6}\ m

Magnetic field B=4\times10^{-8}\ T

Mass of proton m=1.673\times10^{-21}\ Kg

Radius of earth R =6.38\times10^{6}\ m

Radius of orbit r=R+h

r=6.38\times10^{6}+1.790\times10^{6}

r=8170000\ m

We need to calculate the speed

Using formula of magnetic field

Bvq=\dfrac{mv^2}{r}

v=\dfrac{Bqr}{m}

Put the value into the formula

v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}

v=31.25\ m/s

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0
3 years ago
Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0
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Marta_Voda [28]

Answer:

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Explanation:

7 0
2 years ago
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AURORKA [14]
KE = 1/2mv^2
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