Find the acceleration of a hamster when it increases its velocity from rest to 8.0 m/s in 3.0 s .

1 answer:

6 0

<span>Acceleration is the change in velocity divided by time.
We can find acceleration a, by the following formula

a=v-u/t

where,
v is the final velocity (in this question v=8.0 m/s)
u is the initial velocity (since the hamster starts from rest, u=0)
t is the time taken (i,e 3.0 second)
now by applying the formula we have,

a = 8.0 - 0 / 3
= 8 / 3
= 2.65 m/s</span>²<span>

The acceleration is 2.65 meters per second squared</span>

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A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is

valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

$\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2$

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

$\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}$