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4 years ago

Why are scientific theories changed but never thrown away

2 answers:

6 0

All theories have some proof to them and their roots. Although, as science becomes more advanced, we begin to realise that there are some things that we either interpreted wrong, or have since been changed.

Hope this helps! <3

attashe74 [19]4 years ago

4 0

Because still some people believe in them

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Which of the following is a push or pull?

Nutka1998 [239]

Answer: A. Newton

Explanation:

A force is a push or pull exerted on an object and a newton is the type of measurement for a force.

4 0

3 years ago

Read 2 more answers

A ball rolls off a desk at a speed of 2 m/s and lands 0.50 seconds later.

Hoochie [10]

Answer:

(a) The ball lands 1 m ahead of the desk's base.

(b) The height of the desk is 1.225 m.

Explanation:

Given:

Initial velocity of the ball is, $u_{0}=2\ m/s$

Time of flight of the ball is, $t=0.50\ s$

(a)

Let the ball fall at a distance of $x$ from the base of the desk and let the height of the desk be $y$ .

The given motion of the ball is a projectile motion that can be divided into 2 mutually perpendicular directions.

Now, considering only the horizontal motion of the ball. The ball not under the influence of any force in the horizontal direction.

Therefore, the distance covered by the ball horizontally is given as:

$x=v_0\times t\\x=2\times 0.50=1\ m$

So, the ball lands 1 m ahead of the desk's base.

(b)

Now, consider only the vertical motion of the ball. The ball is under the influence of gravity.

So, vertical distance can be obtained using Newton's equations of motion.

We use the following equation of motion:

$y-y_0=u_{oy}t+\frac{1}{2}gt^2$

Where,

$y_{0}\rightarrow \textrm{initial vertical position of ball}\\u_{oy}\rightarrow \textrm{initial vertical velocity}\\g\rightarrow \textrm{acceleration due to gravity}\\y\rightarrow \textrm{final position of ball vertically}$

Here, as the ball leaves the desk horizontally, initial velocity is only in the horizontal direction and doesn't have any component in the vertical direction. So, $u_{oy}=0\ m/s$

Plug in $y_0=h,y=0,g=-9.8,u_{oy}=0,t=0.5$ . Solve for $h$ .

$0-h=0+\frac{1}{2}(-9.8)(0.5)^2\\-h=-4.9\times 0.25\\h=1.225\ m$

Therefore, the height of the desk is 1.225 m.

6 0

3 years ago

A bomb at rest at the origin of an xy-coordinate system explodes into three pieces. Just after the explosion, one piece, of mass

ValentinkaMS [17]

Answer:

Explanation:

a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.

b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.

If v be the velocity of the third part along a direction making angle θ

with x axis ,

x component of v = vcosθ

So momentum along x axis after explosion of third part = mv cosθ

= 10 v cosθ

Momentum along x of first part = - 5 x 42 m/s

momentum of second part along x direction =0

total momentum along x direction before explosion = total momentum along x direction after explosion

0 = - 5 x 42 + 10 v cosθ

v cosθ = 21

Similarly

total momentum along y direction before explosion = total momentum along y direction after explosion

0 = - 5 x 38 + 10 v sinθ

v sinθ= 21

squaring and and then adding the above equation

v² cos²θ +v² sin²θ = 21² +19²

v² = 441 + 361

v = 28.31 m/s

Tanθ = 21 / 19

θ = 48°

6 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

2. A force of 50 N is applied to the object for a distance of 2.0 m. Assume that object（the mass of the object is 3kg）

Gwar [14]

Answer:

2. A force of 50 N is applied to the object for a distance of 2.0 m. Assume that object（the mass of the object is 3kg）

was at rest at the beginning, what speed did it achieved because of the work done on it? (Hint:

Calculate the works performed by the force first.)

I figured that is 8.2m/S，I am just not sure can anyone help me i much appreciate it.

4 0

3 years ago