At which location could you place the south pole of a bar magnet so that it would be pulled toward the magnet shown?

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

4 years ago

Explain why materials such as plastic foam, feathers, and fur are poor conductors of heat.

vampirchik [111]

They are poor conductors because they keep the heat trapped in one spot, while other materials like metal let the heat go through itself.

8 0

3 years ago

A high voltage transmission line with a resistance of 0.51 Ω/km carries a current of 1099 A. The line is at a potential of 1300

Illusion [34]

<h2>Answer:</h2>

$\boxed{P=96.09MW}$

<h2>Explanation:</h2>

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

$R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega$

So we can calculate the power loss as:

$P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}$

Finally, the power loss due to resistance in the line is 96.09MW

5 0

3 years ago

Duplain St. is 300 m long and runs from west to east between Baron and Burkey. If Keith is strolling east from Baron at an avera

Ivahew [28]

Sue from Burkey and Keith from Baron will meet in 2 minutes

Answer: Option b

<u>Explanation:</u>

Time taken can be calculated when distance and the speeds are given. Here speeds of Keith and Sue are given. So, we have to find the relative speeds in order to calculate the time taken.

When two objects travel in same direction the relative speed will be the difference between speeds. Similarly when two objects travel in opposite direction, the relative speed will be the sum of given speeds.

Given:

Speed of Sue from Burkey is 6 km/hr and speed of Keith from Baron is 3 km/hr.

The distance between Burkey and Baron is 300 m.

From the formula, $d=s \times t$

where d is distance,s is speed and t is time

It can be derived that, $t=\frac{d}{s}$

s = sum of given speeds = 3 km/hr + 6 km/hr = 9 km/hr

d = 300 m = 0.3 km $\text {Time }=\frac{0.3 \mathrm{km}}{9 \mathrm{km} / \mathrm{hr}}=\frac{1}{30} h r=2 \text { minutes }$

3 0

3 years ago

A pencil has a density of 0.875g/ml. It has a mass of 3.50 g. What is the volume<br><br>

erica [24]

Answer:

V = 4 cm^3

Explanation:

Divide the mass by the density.

3.50g/0.875g/mL = 4cm^3

7 0

3 years ago