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2 years ago

4cm cubed = ?m cubed

Physics

1 answer:

Neporo4naja [7]2 years ago

5 0

Answer:

${ \rm{1 \: {cm}^{3} = \frac{1}{100 {}^{3} } \: {m}^{3} }} \\ \\ { \rm{4 \: {cm}^{3} = \frac{4}{1000000} \: {m}^{3} }} \\ \\ { \boxed{ \rm{ = 0.000004 \: {m}^{3} }}}$

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When a chemical reaction occurs,

mel-nik [20]

A) reactants interact to form products with different chemical and physical properties

3 0

3 years ago

What was the direction of the ball’s velocity

Tju [1.3M]

Part of the question is missing. Here it is:

<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table. The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity just before it hit the floor? </em>

Answer:

$\theta=-75.7^{\circ}$

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

$s=ut+\frac{1}{2}at^2$

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

t is the time

Solving for t,

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s$

Now we can find the final vertical velocity of the ball, using:

$v_y=u+at$

And susbtituting t = 0.52 s, we find

$v_y = 0 +(9.8)(0.52)=5.1 m/s$

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

$v_x = \frac{0.7}{0.52}=1.3 m/s$

So now we can find the direction of the ball's velocity using:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}$

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

$\theta=-75.7^{\circ}$

8 0

3 years ago

In the past, the intensity of the Earth's magnetic field was _____. A. greater B. weaker C. same D. evidence is inconclusive?

Margarita [4]

I'd say a because it says it's magnetic field as decreased by 10% in the last century

7 0

3 years ago

Read 2 more answers

Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500 m3/s at a location 90 m above the lake s

WINSTONCH [101]

Answer: a) 5 x 10^5 kg/s b) 444 MW

Explanation:

Kinetic energy per unit mass Ke is

Ke = V^2 / 2

Ke = 3^2 / 2 = 4.5 J/kg = 0.0045 kJ/kg

Now potential energy per unit mass Pe is

Pe = g x z = 9.8 x 90 = 882.9 J/kg = 0.8829 kJ/kg

The total mechanical energy of the River per unit mass e = Ke + Pe = 0.0045 + 0.8829 = 0.88744 J/kg

M = P x V = 1000 x 500 = 5 x 10^5 kg/s

b) power generation potential of the entire river at the location Wmax = Emech = M x Emech = 500,000 x 0.88744 = 444,000kW = 444MW

7 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

2 years ago