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3 years ago

FLUKE

Physics

1 answer:

Thepotemich [5.8K]3 years ago

3 0

Answer:

cant understand anything

Explanation:

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. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in

kupik [55]

Answer:

Explanation:

The position is $r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}$ .

The velocity is the first time-derivative of r(t).

$v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}$

The acceleration is the first time-derivative of the velocity.

$a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}$

Since a(t) does not have the variable t, it is constant. Hence, at any time,

$a = -16\hat{j}$

Its magnitude is 16 m/s².

4 0

2 years ago

What is the magnite of the net displacement of the mouse?

agasfer [191]

Complete Question

A field mouse trying to escape a hawk runs east for 5.0m, darts southeast for 3.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the mouse?

Answer:

The values is $s = 7.49 \ m$

Explanation:

From the question we are told that

The distance it travels eastward is $x = 5.0 \ m$

The distance it travel towards the southeast is $l = 3.0\ m$

The distance it travel towards the south is $z = 1 \ m$

Let x-axis be east

y-axis south

z-axis into the ground

The angle made between east and south is $\theta = 45^o$

The displacement toward x-axis is

$x = 5 + 3cos(45)$

$x = 7.12$

The displacement toward the y-axis is

$y = 3 * sin (45)$

$y = 2.123$

Now the overall displacement of the rat is mathematically evaluated as

$s = \sqrt{7.12^2 + 2.12^2 + 1^2}$

$s = 7.49 \ m$

3 0

3 years ago

Help with 5 and the question below. BRAINLIEST FOR THE CORRECT ANSWER! Urgent Physics help!

krok68 [10]

Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

$L = 10 Log\frac{I}{I_0}$

$20 = 10 Log{I}{10^{-12}}$

$I_1 = 10^{-10}W/m^2$

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

$L = 10 Log\frac{I}{I_0}$

$50 = 10Log\frac{I}{10^{-12}}$

$I_2 = 10^{-7}W/m^2$

now we know that

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}$

$r_2 = 47 cm$

so now the distance from friend must be 47 cm

8 0

2 years ago

A series of bright fringes appears on the viewing screen of a Young's double-slit experiment. Suppose you move from one bright f

goblinko [34]

Ahahaha fidjdsd skssjsjsbs SJSU’s

6 0

2 years ago

How much work is done by 0.070 m3 of gas, when the volume remains constant with pressure of 63 x 105 Pa?

stealth61 [152]

Answer:

W = 0 J

Explanation:

The amount of work done by gas at constant pressure is given by the following formula:

$W = P\Delta V$

where,

W = Work done by the gas

P = Pressure of the gas

ΔV = Change in the volume of the gas

Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:

$W = P(0\ m^3)\\$

W = 0 J

5 0

2 years ago