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3 years ago

EASY BRAINLIEST!!URGENT PLEASE HELP.

Physics

1 answer:

Anastaziya [24]3 years ago

8 0

ANSWER:

[i] False.

[ii] Work is the product of force and displacement.

REASONS:

[i] As we already know that temperature is a measure of the average kinetic energy of the particles in an object.

[ii] W = FS

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I'll give brainliest can someone help me with 2.4.1-2.4.3?with explanation.

KIM [24]

Answer:

220÷4=550 per hour

2.4.3 550×4=2200

hope it's right

8 0

3 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

A proton of mass 1.67×10-27 kg is being accelerated along a straight line at 7.01×1015 m/s2 in an accelerator. If the proton has

Masja [62]

Answer:

$v_f=7*10^7\frac{m}{s}$

Explanation:

The proton is under a linear motion with constant acceleration. So, we use the kinemtic equations to calculate its final speed. We know its acceleration, its initial speed and its traveled distance. Thus, we use the following equation:

$v_f^2=v_0^2+2ax\\v_f=\sqrt{v_0^2+2ax}\\v_f=\sqrt{(6.63*10^7\frac{m}{s})^2+2(7.01*10^{15}\frac{m}{s^2})(3.63*10^{-2}m)}\\v_f=7*10^7\frac{m}{s}$

5 0

4 years ago

A storm cloud has a charge of 0.055 C. Due to polarization, the top of a tree gains a charge of -0.006 C. The cloud moves to 98

Anastasy [175]

Answer:

F = 309.24 N

Explanation:

Given that,

Charge on a strom cloud, q₁ = 0.055 C

The charge gained by the top of a tree, q₂ = -0.006 C

The cloud moves to 98 meters above the tree.

We need to find the amount of force between the cloud and the tree. The electrical force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 0.055\times 0.006 }{(98)^2}\\\\F=309.24\ N$

So, the force between the cloud and the tree is equal to 309.24 N.

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3 years ago

What is the speed of a wave frequency of 300hz and a wavelength of 25M

tia_tia [17]

Answer:

7500 m/s

Explanation:

We can use the equation velocity of a wave equals wavelength times frequency. Therefore, v = wavelength*f = (25 m)(300 Hz) = m/s7,500

8 0

3 years ago