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Diano4ka-milaya [45]

3 years ago

8

A car weighing 8000N is traveling at 45 m/s on a perfectly flat, frictionless road. If the driver slams on the brakes, how far w

ill thw car slide before it comes to a stop?

1 answer:

laila [671]3 years ago

3 0

Without friction, the car cannot stop...

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A meter stick A hurtles through space at a speed v = 0.25c relative to you, with its length aligned with the direction of motion

yaroslaw [1]

Answer:

$L_0\approx1.0328\ m$

Explanation:

Given:

relativistic length of stick A, $L=1\ m$

relativistic velocity of stick A with respect to observer, $v=0.25c=7.5\times 10^{7}\ m.s^{-1}$

<em>Since the object is moving with a velocity comparable to the velocity of light with respect to the observer therefore the length will appear shorter according to the theory of relativity.</em>

<u> Mathematical expression of the theory of relativity for length contraction:</u>

$L=\frac{L_0}{\gamma}$

where:

L = relativistic length

$L_0=$ original length at rest

$\gamma =$ Lorentz factor $=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$

$\Rightarrow 1=\frac{L_0}{\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }}$

$L_0=\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }$

$L_0\approx1.0328\ m$

4 0

3 years ago

The space shuttle approaches earth at a speed of 2.89*10^4 km/h when it sends a radio signal to earth 1.00*10^5 km away. The rad

zysi [14]

Answer:

3 secods later

Explanation:

6 0

3 years ago

Does anyone know how to do this question???<br><br> Force = 7kN <br> Pressure = 1mPa<br> Area = ?

Andreas93 [3]

Answer:

7000 m^2

Explanation:

Pressure=Force/Area

1=7000/Area

1(Area)=7000

Area=7000 m^2

5 0

3 years ago

At the end of the adiabatic expansion, the gas fills a new volume V₁, where V₁ > V₀. Find W, the work done by the gas on the

tino4ka555 [31]

Answer:

$W=\frac{p_0V_0-p_1V_1}{\gamma-1}$

Explanation:

An adiabatic process refers to one where there is no exchange of heat.

The equation of state of an adiabatic process is given by,

$pV^{\gamma}=k$

where,

$p$ = pressure

$V$ = volume

$\gamma=\frac{C_p}{C_V}$

$k$ = constant

Therefore, work done by the gas during expansion is,

$W=\int\limits^{V_1}_{V_0} {p} \, dV$

$=k\int\limits^{V_1}_{V_0} {V^{-\gamma}} \, dV$

$=\frac{k}{\gamma -1} (V_0^{1-\gamma}-V_1^{1-\gamma})\\$

(using $pV^{\gamma}=k$ )

$=\frac{p_0V_0-p_1V_1}{\gamma-1}$

4 0

3 years ago

How much work is needed to pump all the water out of a cylindrical tank with a height of 10 m and a radius of 5 m

balu736 [363]

Answer:

Explanation:

volume of water being lifted

= π r² h , where r is radius of cylinder and h is height of cylinder

= 3.14 x5² x 10

= 785 m³

mass of water = 785 x 10³ kg

mass of this much of water is lifted so that its centre of mass is lifted by height

10 / 2 = 5m .

So work done = mgh , m is mass of water , h is displacement of centre of mass and g is acceleration due to gravity

= 785 x 10³ x 9.8 x 5

= 38.465 x 10⁶ J

6 0

4 years ago