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2 years ago

13

Please help!!! These questions are about specific heat capacity.

1 answer:

Elodia [21]2 years ago

4 0

The answer 123=777
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In a fluid-filled container why is the pressure greater at the base of the container

miss Akunina [59]

I believe because there is much less air and much more water in the bottom

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3 years ago

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The theory of plate tectonics suggests that about 200 million years ago, all of Earth's land was confined to a single continent

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3 years ago

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Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m/s^2, where t is in seconds. What

hjlf

To solve this problem we will use the given expression and derive it in order to find the algebraic expressions of velocity and position. These equations will be similar to those already known in the cinematic movement but will be subject to the previously given function. We start deriving the equation for velocity

$a = 2t-6$

$\frac{dv}{dt} = 2t-6$

Integrate acceleration equation

$\int dv = (2t-6)dt$

$v = 2(\frac{t^2}{2})-6t+C_1$

$v=t^2-6t+C_1$

At $t = 0, v = 0$

Replacing,

$0 = 0^2-6*0+C_1$

Therefore the value of the first Constant is

$C_1 = 0$

The expression can be escribed as,

$v = t^2-6t$

Calculate the velocity after 6s,

$v=t^2-6t$

$v = 6^2-6*6$

$v = 0m/s$

Now using the same expression we can derive the equation for distance

$v = t^2-6t$

$\frac{dx}{dt} =t^2-6t$

$\int dx = \int (t^2-6t)dt$

$x = \frac{t^3}{3}-6\frac{t^2}{2}+C_2$

At t=0, x=0

$0 = \frac{0^3}{3}-6(\frac{0^2}{2})+C_2$

Therefore the value of the second constant is

$C_2 = 0$

$x = \frac{t^3}{3}-\frac{6t^2}{2}$

Calculate the distance traveled after 11 s

At $t=11s$

$x = \frac{11^3}{3}-6(\frac{11^2}{2})$

$x = 80.667m$

6 0

3 years ago

"Angel Falls in southeastern Venezuela is the highest uninterrupted waterfall in the world, dropping 979 m (3212 ft). Ignoring a

Furkat [3]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $v = 3.57 \ m/s$

Explanation:

From the question we are told that

The depth is $d = 979 \ m$

The time taken for water to fall to below the river is $t = 14 \ s$

The horizontal distance from the base of the vertical cliff is $h = 50 \ m$

Generally the horizontal speed at the top is mathematically represented as

$v = \frac{h}{t}$

=> $v = \frac{ 50}{ 14 }$

=> $v = 3.57 \ m/s$

4 0

2 years ago

A new spacecraft is built with an engine that is able to produce constant acceleration of 1g, where g = gravity’s acceleration o

QveST [7]

given that acceleration due to gravity is g = 10 m/s^2

speed of the rocket will reach to 0.9c

$v_f = 0.9* 3 * 10^8 = 2.7 * 10^8 m/s$

now by kinematics

$v_f = v_i + a*t$

$2.7 * 10^8 = 0 + 10*t$

$t = 2.7*10^7 s$

Part b)

distance traveled by it in above time

$d = v*t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}*10*(2.7*10^7)^2$

$d = 3.645 * 10^{15} m$

so this is the distance covered by the object

3 0

3 years ago