The collision of two plates causes the formation of

The specific heat capacity of sea water is 4100 J/Kg°C and the boiling point of 100.6 °C. (i) Calculate the energy required to r

maw [93]

Answer:

334.314 (kJ)

Explanation:

1) the formula for the required energy is: Q=c*m(Bp-t), where c - 4100 J/kg*C; m - 0.9 kg; Bp - 100.6 C; t - 10 C.

2) according to the formula above:

Q=4100*0.9*(100.6-10)=41*9*906=334314 (J).

6 0

3 years ago

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0

4 years ago

What is the period of a pendulum that is 1.3 m long?

Mkey [24]

Answer:

T≈2.2876585681s

Explanation:

Since you were not given the degrees of amplitude or radians of amplitude, we assume that the degrees of amplitude are less than 20∘ or 20∘ and we also assume that this is a simple pendulum

6 0

3 years ago

Find the distance between the points given. (-3, 2) and (9, -3) 13

ohaa [14]

The distance between two points on a graph is

Square root of [ (difference in the y-values)² + (difference in the x-values)² ] .

So for these two points, the distance between them is

square root of [ (-5)² + (12)² ] =

square root of [ 25 + 144 ] =

square root of [ 169 ] =

Yes, the ' 13 ' you typed at the end of the question is the correct answer to it.

5 0

3 years ago

I would like to know why this is the correct answer

Helen [10]

The acceleration of the object if the net force is decreased = 0.13 m/s²

<h3>Further explanation</h3>

Given

A net force of 0.8 N acting on a 1.5-kg mass.

The net force is decreased to 0.2 N

Required

The acceleration of the object if the net force is decreased

Solution

Newton's 2nd law :

$\tt \sum F=m.a$

The mass used in state 1 and 2 remains the same, at 1.5 kg

state 1

ΣF=0.8 N

m=1.5 kg

The acceleration, a:

$\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{0.8}{1.5}\\\\a=0.53`m/s^2$

state 2

ΣF=0.2 N

m=1.5 kg

The acceleration, a:

$\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{0.2}{1.5}\\\\a=0.13~m/s^2$

8 0

3 years ago