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3 years ago

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I need help, please help an object starts from rest at time t=0 and moves in the with constant acceleration. The object travels

3m from time t=1 to time t=2. what is the acceleration of the object?

1 answer:

vichka [17]3 years ago

4 0

<span>a=8(m/<span>sec2</span>).. i answered this earlier i think.</span>

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Which of the following are in the correct order from smallest or largest?

maks197457 [2]

B) millimeter, centimeter, meter, kilometer

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2 years ago

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If its wavelength were doubled, its energy would be If its wavelength were doubled, its energy would be 4E. 2E. 12E. 14E.

ANEK [815]

The wavelength was doubled, and its energy will be increased by 4 times.

looking at the formula

energy $E = MC^2$

also, $c = \lambda \times \nu$

hence it is clear from above that energy is directly proportional to the square of the wavelength.

hence, The wavelength was doubled, and its energy will be increased by 4 times.

<h3>What is Wavelength?</h3>

The distance over which a periodic wave's shape repeats is known as the wavelength in physics.
It is a property of both traveling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings.
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brainly.com/question/13533093

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1 year ago

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

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3 years ago

Which reading strategy helps you to read and understand a passage quickly by making use of your previous knowledge?

Molodets [167]

Taking notes,because you can read back on what you learned

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3 years ago

Three masses are located in the x- y plane as follows: a mass of 6 kg is at (0 m, 0 m), a mass of 4 kg is at (3 m, 0 m), and a m

enot [183]

Answer:

The center of mass of three mass in the x-y plane is located at (1,0.5).

Explanation:

It is given that, a mass of 6 kg is at (0,0), a mass of 4 kg is at (3,0), and a mass of 2 kg is at (0,3). We need to find the center of mass of the system. Center of mass in x direction is :

$C_x=\dfrac{6\times 0+4\times 3+2\times 0}{6+4+2}\\\\C_x=1$

The center of mass in y direction is :

$C_y=\dfrac{6\times 0+4\times 0+2\times 3}{6+4+2}\\\\C_y=0.5$

So, the center of mass of three mass in the x-y plane is located at (1,0.5).

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3 years ago