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3 years ago

7. A car travels at 25 m/s to the North. It has an acceleration of 2 m/s to the south

Physics

1 answer:

tamaranim1 [39]3 years ago

7 0

Answer:

-0.5 m/s2 as it comes to rest.

Explanation:

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Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

3 0

3 years ago

A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of

maks197457 [2]

Answer:

$5.6\times 10^{-8}\ Ohm.m$

Explanation:

Resistivity is given by $\rho=\frac {AR}{L}$ where A is cross-sectional area, R is resistance, L is the length and $\rho$ is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

$\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m$

8 0

3 years ago

A jet is circling an airport control tower at a distance of 20.6 km. An observer in the tower watches the jet cross in front of

lesya [120]

Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = $20.6\times 10^3\ m$

$\theta$ = The angle subtended by moon = $9.6\times 10^{-3}\ rad$

Distance traveled is given by

$s=r\times\theta$

$\Rightarrow s=20.6\times 10^3\times 9.6\times 10^{-3}$

$\Rightarrow s=197.76\ m$

The distance traveled by the jet is 197.76 m

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3 years ago

A less massive moving object has an elastic collision with a more massive object that is not moving. Compare the initial velocit

stepladder [879]

Assume that the small-massed particle is $m$ and the heavier mass particle is $M$ .

Now, by momentum conservation and energy conservation:

$mv = mv_{m} + Mv_{M}$

$mv^{2} = mv^{2}_{m} + Mv^{2}_{M}$

Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

$v_{m} = v\frac{m - M}{m + M}\\\\\\v_{M} = v\frac{m + M}{m + M}\\$