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2 years ago

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Calculate the average distances the car and the washer traveled from the top of the track. Record the average in Table A of your

Student Guide. What is the average distance the car traveled from the top of the track? cm What is the average distance the washer traveled from the top of the track? cm

2 answers:

Romashka-Z-Leto [24]2 years ago

7 0

The answer for both is 247 cm

KIM [24]2 years ago

7 0

Both is 247. hope it helps.

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There are Z protons in the nucleus of an atom, where Z is the atomic number of the element. An α particle carries a charge of +2

qaws [65]

Answer:

$r = 2.84 \times 10^{-14} m$

Explanation:

As per energy conservation we know that the electrostatic potential energy of the charge system is equal to the initial kinetic energy of the alpha particle

So here we can write it as

$\frac{1}{2}mv^2 = \frac{k(2e)(ze)}{r}$

now we know that

$m = 1.67 \times 10^{-27} kg$

$e = 1.6 \times 10^{-19} C$

z = 79

here kinetic energy of the incident alpha particle is given as

$KE = 6.4 \times 10^{-13} J$

now we have

$6.4 \times 10^{-13} = \frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r}$

now we have

$r = 2.84 \times 10^{-14} m$

7 0

3 years ago

A motorcycle is stopped at a traffic light. When the light turns green, the motorcycle accelerates to a speed of 91 km/h over a

zimovet [89]

Given :

Initial speed , u = 0 m/s .

Final speed , v = 91 km/h = 25.28 m/s .

To Find :

a) Average acceleration .

b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .

Solution :

a )

We know ,by equation of motion :

$v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{25.28^2-0^2}{2\times 47}\ m/s^2\\\\a=6.8\ m/s^2$

b)

Also , by equation of motion :

$s=ut+\dfrac{at^2}{2}\\\\s=0+\dfrac{6.8\times (3.3)^2}{2}\ m\\\\s=37.02\ m$

Hence , this is the required solution .

6 0

3 years ago

A thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/

Artyom0805 [142]

Answer:

$I=182.97\ kg-m^2$

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, $\tau=860\ N$

Angular acceleration of the shell, $\alpha =4.7\ m/s^2$

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

$\tau=I\alpha$

I is the rotational inertia of the shell

$I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2$

So, the rotational inertia of the shell is $182.97\ kg-m^2$ .

7 0

3 years ago

PLEASE ITS AN Emergency IF ITS RIGHT I WILL GIVE BRAINLIEST

Art [367]

Uh I think it’s point guard

3 0

2 years ago

A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holdi

Elis [28]

Answer: $115.52\ N$

Explanation:

Given

Length of plank is 1.6 m

Force $F_1$ is applied on the left side of plank

Force $F_2$ is applied 43 cm from the left end O.

Mass of the plank is $m=13.7\ kg$

for equilibrium

Net torque must be zero. Taking torque about left side of the plank

$\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N$

Net vertical force must be zero on the plank

$\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N$

8 0

2 years ago