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3 years ago

If a cannon is fired use a=F/m to explain what happens to the cannon and the cannonball.

Physics

2 answers:

Serga [27]3 years ago

4 0

Im not really sure about this im stuck to

Harlamova29_29 [7]3 years ago

3 0

The cannon and the cannonball both blow up or shatter to little peices

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Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

A 320 g air track cart traveling at 1.25 m/s collides with a stationary 270 g cart. What is the speed of the 270 g cart after th

Nutka1998 [239]

Answer:

The speed of the 270g cart after the collision is 0.68m/s

Explanation:

Mass of air track cart (m1) = 320g

Initial velocity (u1) = 1.25m/s

Mass of stationary cart (m2) = 270g

Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s

7 0

3 years ago

What is the acceleration of a 68 kg block of

S_A_V [24]

Answer:

Explanation:

Formula of acceleration a = fm

A = 68 x 386

= 26248

8 0

3 years ago

How does a wind turbine make electricity?

Marysya12 [62]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1

Nonamiya [84]

Answer:

$a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

Magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$

Explanation:

$t_1=2\ \text{s}$

$v_x=1\ \text{m/s}$

$v_y=3\ \text{m/s}$

$t_2=2.5\ \text{s}$

$v_x=4\ \text{m/s}$

$v_y=3\ \text{m/s}$

Average acceleration in the different axes

$a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2$

$a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2$

The components of the acceleration is $a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

The magnitude of acceleration

$a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2$

Direction

$\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}$

The magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$ .

7 0

2 years ago