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3 years ago

How do you find the angle (to the nearest tenth) on your graph that produced the greatest range, based on the line of best fit?

Physics

1 answer:

VikaD [51]3 years ago

3 0

Answer:

yuououiy

Explanation:

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Ok plz help this is my entrance exam question for physics in grade 9 (yes i am moving to a new school again)

vagabundo [1.1K]

Answer:

300 W

Explanation:

We know that

P = W/t

P = mgh/t

P = 40 × 10 × 3/4

P = 10 × 10 × 3

P = 300 W

4 0

3 years ago

A fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative

Liula [17]

Question is missing. Found on google:

a) What are the components of the acceleration of the fish?

(b) What is the direction of its acceleration with respect to unit vector î?

(a) $(0.73, -0.47) m/s^2$

The initial velocity of the fish is

$u=(4.00 i + 1.00 j) m/s$

while the final velocity is

$v=(15.0 i - 6.00 j) m/s$

Initial and final velocity are related by the following suvat equation:

$v=u+at$

where

a is the acceleration

t is the time

The time in this case is t = 15.0 s, so we can use the previous equation to find the acceleration, separating the components:

$v_x = u_x + a_x t\\a_x = \frac{v_x-u_x}{t}=\frac{15.0-4.00}{15.0}=0.73 m/s^2$

$v_y = u_y + a_y t\\a_y = \frac{v_y-u_y}{t}=\frac{-6.00-1.00}{15.0}=-0.47 m/s^2$

(b) $-32.8^{\circ}$

The direction of the acceleration vector with respect to i can be found by using the formula

$\theta = tan^{-1}(\frac{a_y}{a_x})$

where

$a_x$ is the horizontal component of the acceleration

$a_y$ is the vertical component of the acceleration

From part a), we have

$a_x = 0.73 m/s^2$

$a_y = -0.47 m/s^2$

Substituting,

$\theta = tan^{-1}(\frac{-0.47}{0.73})=-32.8^{\circ}$

(c) $r=(460.5 i - 185.1 j )m$

The initial position of the fish is

$r_0 = (12.0 i -3.60 j) m$

The generic position r at time t is given by

$r= r_0 + ut + \frac{1}{2}at^2$

where

$u=(4.00 i + 1.00 j) m/s$ is the initial velocity

$a=(0.73 i -0.47 j) m/s^2$ is the acceleration

Substituting t = 30.0 s, we find the final position of the fish. Separating each component:

$r_x =12.0 + (4.00)(30) + \frac{1}{2}(0.73)(30)^2=460.5 m\\r_y = -3.60 + (1.00)(30) + \frac{1}{2}(-0.47)(30)^2=-185.1 m$

So the final position is

$r=(460.5 i - 185.1 j )m$

4 0

3 years ago

Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water. If the cylinder

jonny [76]

Answer:

0.10839 m

Explanation:

$P_1$ = Atmospheric pressure = 1 atm = 101325 Pa

$P$ = Total pressure at bottom of mecury = 1.2 atm

g = Acceleration due to gravity = 9.81 m/s²

h = d = Depth of mercury

$\rho_m$ = Density of mercury = $1.36\times 10^4\ kg/m^3$

$\rho_w$ = Density of water = $1000\ kg/m^3$

Pressure at the bottom is of the cylinder is given by

$P_2=P_1+\rho_wgh\\\Rightarrow P_2=101325+1000\times 9.81(0.7-d)$

Pressure at the bottom of mercury is

$P=P_2+\rho_mgh\\\Rightarrow 1.2\times 101325=101325+1000\times 9.81(0.7-d)+1.36\times 10^4\times 9.81\times d\\\Rightarrow 1.2\times 101325=123606d+108192\\\Rightarrow d=\dfrac{1.2\times 101325-108192}{123606}\\\Rightarrow d=0.10839\ m$

The depth of the mercury is 0.10839 m

7 0

4 years ago

The charge entering the positive terminal of an element is q = 5 sin 4πt mC while the voltage across the element (plus to minus)

Artemon [7]

Answer:

(a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

Explanation:

Given that,

Charge $q=5\sin4\pi t\ mC$

Voltage $v=3\cos4\pi t\ V$

Time t = 0.3 sec

We need to calculate the current

Using formula of current

$i(t)=\dfrac{dq}{dt}$

Put the value of charge

$i(t)=\dfrac{d}{dt}(5\sin4\pi t)$

$i(t)=5\times4\pi\cos4\pi t$

$i(t)=20\pi\cos4\pi t$

(a).We need to calculate the power delivered to the element

Using formula of power

$p(t)=v(t)\times i(t)$

Put the value into the formula

$p(t)=3\cos4\pi t\times20\pi\cos4\pi t$

$p(t)=60\pi\times10^{-3}\cos^2(4\pi t)$

$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi t}{2})$

Put the value of t

$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi\times0.3}{2})$

$p(t)=30\pi\times10^{-3}(1+\cos8\pi \times0.3)$

$p(t)=187.68\ mW$

(b). We need to calculate the energy delivered to the element between 0 and 0.6 s

Using formula of energy

$E(t)=\int_{0}^{t}{p(t)dt}$

Put the value into the formula

$E(t)=\int_{0}^{0.6}{30\pi\times10^{-3}(1+\cos8\pi \times t)}$

$E(t)=30\pi\times10^{-3}\int_{0}^{0.6}{1+\cos8\pi \times t}$

$E(t)=30\pi\times10^{-3}(t+\dfrac{\sin8\pi t}{8\pi})_{0}^{0.6}$

$E(t)=30\pi\times10^{-3}(0.6+\dfrac{\sin8\pi\times0.6}{8\pi}-0-0)$

$E(t)=57.52\ mJ$

Hence, (a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

6 0

3 years ago

A car slows down at -5.00 m/s^2 until it comes to a stop after travelling 15.0 m. How much time did it take to stop?

Xelga [282]

In physics, there are already derived equation that are based on Newton's Law of Motions. The rectilinear motions at constant acceleration have the following equations:

x = v₁t + 1/2 at²
a = (v₂-v₁)/t

where
x is the distance travelled
v₁ is the initial velocity
v₂ is the final velocity
a is the acceleration
t is the time

Now, we solve first the second equation. Since it mentions that the car comes eventually to a stop, v₂ = 0. Then,

-5 = (0-v₁)/t
-5t = -v₁
v₁ = 5t

We use this new equation to substitute to the first one:
x = v₁t + 1/2 at²
15 = 5t(t) + 1/2(-5)t²
15 = 5t² - 5/2 t²
15 = 5/2 t²
5t² = 30
t² = 30/5 = 6
t = √6 = 2.45

Therefore, the time it took to travel 15 m at a deceleration of -5 m/s² is 2.45 seconds.

5 0

3 years ago