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qwelly [4]
3 years ago
15

to start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 53.0° above the ho

rizontal. it explodes on the mountainside 35.0 s after firing. what are the x and y coordinates of the shell where it explodes, relative to its firing point?
Physics
1 answer:
kap26 [50]3 years ago
5 0
So this is easy to calculate when you split the velocity into x and y components. The x component is going to equal cos(53) * 290 and the y component is going to equal sin(53)*290.

The x location therefore is 290*cos(53)*35 = 6108.4m

The y location needs to factor in the downwards acceleration of gravity too, which is 9.81m/s^2. We need the equation dist. = V initial*time + 0.5*acceleration*time^2.

This gives us d=290*sin(53)*35 + (0.5*-9.81*35^2)=2097.5m

So your (x,y) coordinates equals (6108.4, 2097.5)
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3 years ago
Click the Run Now button to start the simulations. Select "Many rays" and click the Screen checkbox. You should see a lamp and a
Gnoma [55]

Answer:

Answer explained below

Explanation:

(a) The rays are diverging near the lens. They change the direction when they passed through the converging lens

(b) If the light rays don't bend they will move away from the optical (principal axis) as the other waves are moving.

(c) If we decrease the distance between lens and light source, most of the rays diverge and no ray converges on the screen even after passing through the lens. Here is a screenshot.

5 0
3 years ago
Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec
sammy [17]

Answer:

1) The maximum jump height is reached at A. 0.337s

2) The maximum center of mass height off of the ground is B. 1.64m

3) The time of flight is C. 0.834s

4) The distance of jump is B. 7.49m

Explanation:

First of all we need to decompose velocity in its rectangular components, so

v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s

1) We use, v_{fy}=v_{iy}-gt, as we clear it for t and using the fact that v_{fy}=0 at max height, we obtain t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s

2) We can use the formula y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2} for t=0.337s, so

y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m

3) We can use the formula y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find total time of fligth, so 0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66, as it is a second-grade polynomial, we find that its positive root is t=0.834s

4) Finally, we use x=v_{x}t=8.05m/s(0.834s)=6.71m, as it has an additional displacement of 0.77m due the leg extension we obtain,

x=6.71m+0.77m=7.48m, aprox x=7.49m

5 0
3 years ago
A piano string having a mass per unit length of 5.00 g/m is under a tension of 1350 N. Determine the speed of transverse waves i
padilas [110]

Answer:

The speed of transverse waves in this string is 519.61 m/s.

Explanation:

Given that,

Mass per unit length = 5.00 g/m

Tension = 1350 N

We need to calculate the speed of transverse waves in this string

Using formula of speed of the transverse waves

v=\sqrt{\dfrac{T}{\mu}}

Where, \mu = mass per unit length

T = tension

Put the value into the formula

v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}

v =519.61\ m/s

Hence, The speed of transverse waves in this string is 519.61 m/s.

6 0
3 years ago
Two families meet at the park at 10:00am. Each family uses a different way of transportation to get there. The Gonzalez family l
pav-90 [236]

Firstly, we need to make the times and distance equal to compare.

1hr=60mins
Gonz drove 90km in 60mins,so they would drive 45km in 30 mins dividing it by 2.
Because the Rivs had gone 30km in 30mins, Gonz were faster.

For the adverage speed, we should first add the speed of each family in mph , so 90+60, which equals 150 and divide that by 2 because there are 2 speeds so the average speed is 75mph.

P.S English is my second language so tell me if you don't understand something and I will try and explain.
5 0
3 years ago
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