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3 years ago

The wetter the surface, the friction

Physics

1 answer:

allochka39001 [22]3 years ago

7 0

Answer: Water can either increase or decrease the friction between surfaces.

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Over a span of 6.0 seconds, a car changes it's speed from 89 km/h to 37 km/h. What is its average acceleration in meters per sec

scoundrel [369]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) - (speed at the beginning)

change in speed = (37 km/hr) - (89 km/hr) = -52 km/hr

Acceleration = (-52 km/hr) / (6 sec)

Acceleration = (-26/3) km/(hr·sec)

Units: (1/hr·sec) · (hr/3600 sec) = 1 / 3600 sec²

(-26/3) km/(hr·sec) = (-26/3) km/(3600 sec²)

= -26,000/(3 · 3600) m/s²

<em>Acceleration = -2.41 m/s²</em>

3 0

3 years ago

The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,

kondaur [170]

Answer:

$a= -2\ m/s^2$

$a=-12.5\ m/s^2$

x=2.17 m

x=8.4 m

Explanation:

Given that

$v=A+Bt^{-1}$

$v=2+0.25t^{-1}$

To find acceleration :

we know that

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=0-0.5t^{-2}$

$a=-0.5t^{-2}$

Acceleration at t= 2 s

$a=-0.5\times 2^{-2}$

$a= -2\ m/s^2$

Acceleration at t= 5 s

$a=-0.5\times 5^{-2}$

$a=-12.5\ m/s^2$

We know that

$v=\dfrac{dx}{dt}$

$dx=\left(2+\dfrac{1}{4t}\right)dt$

Position at t= 2 s:

$\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{2}$

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

$\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{5}$

x=8+0.25 ln5

x=8.4 m

4 0

2 years ago

When exposed to a radioactive source which emits 1.2-MeV gamma-rays, a particular material is found to have a half-value thickne

iren2701 [21]

Answer:E

Explanation:

It is given that Energy of gamma ray is E=1.2 Mev

Shielding effect can be measured by measuring the fraction of gamma rays blocked by shield. If certain thickness will able to block half the radiation then to block 75% radiation we need to add same amount of thickness in order to block the remaining radiation.

i.e. $\frac{E}{2}$ fraction is blocked by 10 cm thickness