A heating element has a resistance of 180 Ω and draws a current of 1.60 A. What is its power?

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0km/hour is blowing southward.

qaws [65]

Answer:

a) correct answer is C , b) 14º from the west to the north, c) v_{1g} = 300.79 km / h

Explanation:

This is a relative speed exercise using the addition of speeds.

1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C

2 and 3) In this case we must compose the speed using the Pythagorean Theorem.

$v_{1a}$ ² = $v_{1g}$ ² + $v_{ag}$ ²

where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth

in this case let's clear the speed of the airplane with respect to the Earth

v_{1g} = √(v_{1a}² - v_{ag}²)

v_{1g} = √ (310² - 75²)

v_{1g} = 300.79 km / h

we find the direction of the airplane using trigonometry

sin θ = v_{ag} / v_{1a}

θ = sin⁻¹ (v_{ag} /v_{1a})

θ = sin⁻¹ (75/310)

θ= 14º

the pilot must direct the aircraft at an angle of 14º from the west to the north

7 0

3 years ago

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .

Plugging this value in

$t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}$

$t=1.06\ seconds$

So, the time of flight of the ball is 1.06 seconds.

6 0

4 years ago

The slope of the line on a speed-time graph tells the speed.

iren2701 [21]

From the choices provided, the better answer is ' T ' .

8 0

3 years ago

A bicycle pump contains 20 cm3 of air at a pressure of 100 kPa. The air is then pumped in a tyre of volume 100 cm3. Calculate th

Natasha2012 [34]

Answer:

The pressure of the air in the tyre is 20 kPa

Explanation:

The parameters for the bicycle pump and tyre are;

The volume of air contained in the bicycle pump, V₁ = 20 cm³

The pressure of the air contained in the bicycle pump, P₁ = 100 kPa

The volume (available) of the tyre, where the air is pumped, V₂ = 100 cm³

Let P₂ represent the pressure in the tyre after the air is pumped

By Boyle's law, we have that at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure;

Mathematically, Boyle's law gives the following equation;

P₁ × V₁ = P₂ × V₂

∴ P₂ = (P₁ × V₁)/V₂

Substituting the known values gives;

P₂ = (100 kPa × 20 cm³)/(100 cm³)

∴ P₂ = 100 kPa × 1/5 = 20 kPa

P₂ = 20 kPa

The pressure of the air in the tyre = P₂ = 20 kPa.

7 0

3 years ago

A car has uniformly accelerated from rest to a speed of 25m/s after traveling 75m. What is its acceleration in m/s^2

Roman55 [17]

<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’): </em></h2><h2><em> </em></h2><h2><em>The car’s speed is 25m/s </em></h2><h2><em>The distance travelled is 75m </em></h2><h2><em>Then we have the formulas for speed and distance: </em></h2><h2><em> </em></h2><h2><em>v = a x t -> 25 = a x t </em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2 </em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say: </em></h2><h2><em> </em></h2><h2><em>t = 25 / a </em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a </em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1): </em></h2><h2><em> </em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2 </em></h2><h2><em>t^2 = 150 / a </em></h2><h2><em>Since t has the same value for both truths we can say: </em></h2><h2><em> </em></h2><h2><em>625 / a^2 = 150 / a </em></h2><h2><em> </em></h2><h2><em>Thus multiply both sides with a^2: </em></h2><h2><em> </em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17 </em></h2><h2><em> </em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>

4 0

3 years ago