All categories

4 years ago

How to find average acceleration?

Physics

1 answer:

Elanso [62]4 years ago

3 0

Answer:

Explanation:

A way to see this is that the definite integral of the acceleration is the change in velocity (i.e. the final velocity minus the initial velocity), and the change in velocity divided by the length of the time interval is the average acceleration on the interval.

You might be interested in

What structure on the south african coast has a range of 63 km and releases flashes every 30 seconds?

I am Lyosha [343]

Answer:

Cape Point lighthouse

Explanation:

The lighthouse is located South of Cape Town, at a point where boats turn to go around the tip of Africa.

The new lighthouse has the most powerful light of all the lighthouses in South Africa, with a power of 10 megacandelas. That's why it's being visible for up to 63 km around.

It's a very important lighthouse helping to conduct the traffic around dangerous waters where 2 oceans meet.

7 0

4 years ago

The distance between two parallel wires carrying currents of 10 A and 20 A is 10 cm. Determine the magnitude and direction of th

Marianna [84]

Explanation:

It is given that,

Current in wire 1, I₁ = 10 A

Current in wire 2, I₂ = 20 A

Distance between wires, d = 10 cm = 0.1 m

Force per unit length is given by :

$\dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}$

$\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 10\times 20}{2\pi \times 0.1}$

$\dfrac{F}{l}=0.0004\ N/m$

$\dfrac{F}{l}=4\times 10^{-4}\ N/m$

So, the magnetic force acting per unit length of the wires $4\times 10^{-4}\ N/m$ . Since, the current is in same direction. So, the force is attractive in nature.

6 0

3 years ago

Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding the hose in herhan

sweet [91]

Answer:

84.196%

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$y_0=1\ m$

x = 10 m

t = Time taken

$v_0$ = 3.5 m/s (assumed, as it is not given)

$A_0$ = $\pi 1.5^2$

We have the equation

$y=y_0+ut+\dfrac{1}{2}gt^2\\\Rightarrow 0=y_0-\dfrac{1}{2}gt^2\\\Rightarrow t=\sqrt{\dfrac{2y_0}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1}{9.81}}\\\Rightarrow t=0.45152\ s$