All categories

4 years ago

Question: Is a flame matter?

Physics

1 answer:

Butoxors [25]4 years ago

5 0

Answer:

Depends on the point you want to make,if you are making the point a flame is not matter than you will put something like....

Flames are not matter because(a point that you have saying it isnt ).

If you are saying it is then put something like.... flames are matter because (a point you have that says it is.)

Explanation:

You might be interested in

A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 73° above the hor

Brrunno [24]

Answer:

7 s

Explanation:

u = Initial velocity of rock = 12.2 m/s

$\theta$ = Angle of throw = $73^{\circ}$

x = Displacement in x direction = 25 m

Displacement in x direction is given by

$x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{25}{12.2\times \cos73^{\circ}}\\\Rightarrow t=7\ \text{s}$

Time taken to reach the ground is 7 s.

4 0

3 years ago

This is a cell, which is the basic unit of all life. All organs in human bodies are made of cells and require oxygen to survive.

kotegsom [21]

The two systems that work together to deliver oxygen are D, respiratory and cardiovascular

5 0

3 years ago

A test charge of +4 µC is placed halfway between a charge of +6 µC and another of +2 µC separated by 20 cm. (a) What is the magn

S_A_V [24]

Answer:

(a) Magnitude: 14.4 N

(b) Away from the +6 µC charge

Explanation:

As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:

$F_e = K\frac{qq_{test}}{r^2}$

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.

Let's say that a force that goes toward the +6 µC charge is positive, then:

$F_e_1 = K\frac{q_1q_{test}}{r^2}=-9*10^9 \frac{Nm^2}{C^2} \frac{6*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =-21.6 N$

$F_e_2 = K\frac{q_2q_{test}}{r^2}=9*10^9 \frac{Nm^2}{C^2} \frac{2*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =7.2 N$

The magnitude will be:

$F_e = -21.6 + 7.2 = -14.4 N$ , away from the +6 µC charge

3 0

4 years ago

Using the projectiles equations for y and x at what second does the ball have a vertical velocity of -39.2 m/s^2

vovangra [49]

Answer:

Explanation:

In projectile the time taken by an object to reach its maximum height is the maximum time (tmax).

tmax = Usin $\theta$ /g

U = velocity of the ball

$\theta$ = angle of projection = 90° (vertical velocity)

g = acceleration due to gravity = 9.8m/s²

tmax = Usin90°/9.8

tmax = U/9.8

T the maximum height U = Uy = 39.2 m/s²

tmax = 39.2/9.8

tmax = 4secs

To calculate the time at which the ball will have an horizontal velocity, we will use the same formula but $\theta$ will be 0° in this case

If Ux = Ucos $\theta$

8 = Ucos0

U = 8m/s

Since t = Usin $\theta$ /g

t = 8sin0/9.8

t = 0sec

8 0

3 years ago

The mass of a basketball is three times greater then the mass of a softball. Compare the momentum’s of a softball and a basketba

vagabundo [1.1K]

Let m = the mass of the softball.
Then the mass of the basketball is 3m.

By definition, momentum = mass * velocity.
If both the softball and basketball move at the same velocity, v, then
Momentum of the softball = mv
Momentum of the basketball = 3mv

Answer: The momentum of the basketball is 3 times that of the softball.

5 0

4 years ago