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3 years ago

PLEASE HELP ASAP!!!!

Physics

1 answer:

Zolol [24]3 years ago

4 0

T1 = 2×W / squrt (3) +1

T2 = squrt(6) ×W/ squrt (3) +1

Here, T1 is along A.

T2 is along B

W = 250N.

Yo can directly use above formula to find solution.

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3. Write the numbers given below in scientific notation.

KIM [24]

Answer:
a) 3 * 10^9 ms
b) 1.6 * 10^-9 gm
c) 6.4 * 10^6 m
d) 5.48 * 10^5 s
Explanation:
Scientific notation is always written in [number from 0-10] (multiplied by) [10] raised to the power 'n'

4 0

2 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

larisa86 [58]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

Mass of the larger disk = $M_2\ =\ 1.60\ kg$
Mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the larger disk = $R_2\ =\ 5.00\ cm\ =\ 0.05\ m$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$
Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

$\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

$mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)$

From the pulley,

$\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)$

From the equation (1) and (2),

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2$

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = $R_2\ =\ 0.05\ m$

Let $a_2$ be the acceleration of the block while connecting to the larger pulley. $\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2$

4 0

3 years ago

For saving lives, what is the most important safely feature on a car

Lubov Fominskaja [6]

Safety belt airbags can kill anyone 12 and under up front plus how u gonna get them in the back

7 0

4 years ago

Read 2 more answers

You build a solenoid containing 500 windings over a 0.20-m length, with a loop radius of 0.025 m for each winding. the current i

marta [7]

The inductance of a solenoid is given by:
$L= \frac{\mu_0 N^2 A}{l}$
where
$\mu_0$ is the vacuum permeability
N is the number of turns of the solenoid
A is the area of one loop of the solenoid
l is its length

For the solenoid in our problem, N=500, and the radius of one loop is r=0.025 m, so the area of one loop is
$A=\pi r^2 = \pi (0.025 m)^2 = 1.96 \cdot 10^{-3} m^2$
and its length is l=0.20 m

So the inductance of the solenoid is
$L= \frac{(4 \pi \cdot 10^{-7} NA^{-2})(500)^2(1.96 \cdot 10^{-3} m^2)}{ 0.20 m}=3.08 \cdot 10^{-3} H$

4 0

4 years ago

if you wanted to increase the energy of a wave but you had to leave the wavelength the same what might you do

Blizzard [7]

Answer:

Increase the amplitude

Explanation:

The energy conveyed by a wave is directly proportional to the square of its amplitude. Thus; E ∝ A²

This means that increasing the amplitude will lead to an increase in the energy.

Now, the amplitude of a wave is the height of a wave from it's highest point known as the peak, to the lowest point on the wave known as the trough whereas wavelength refers to the length of a wave from one peak to the next.

This means that increasing the amplitude has no effect on the wavelength.

7 0

3 years ago