Can you please help me with these physics displacement questions? If you can only do one, that is okay

Physics

1 answer:

Brut [27]3 years ago

7 0

#3). Your drawing in the lower right corner is correct. You're headed down the right road, but ran out of gas and just stopped.

Radius of the circle = 1.5 km

Circumference of the whole circle = (2·π·radius) = 9.42 km

Distance = 3/4 of the way around it = 7.07 km .

Displacement = the straight line from the West point to the North point. The straight-line length is 2.12 km; the straight-line direction from start to finish is Northeast (45°). I'll let you figure out why these numbers.

#4). What if you walk 1 mile East and then 1 mile West ? You got a good workout, and you're back home where you started ! Your distance is 2 miles, and your displacement is zero.

The whale had a good workout too. She swam (6.9 + 1.8 + 3.7) = 12.4 km. She's sweating and tired. Her total distance during that workout is 12.4 km.

Her displacement is the line from start-point to end-point. How she got there doesn't matter, so swimming 1 km East and then swimming 1 km West cancel out, and have no effect on the displacement.

(6.9E + 1.8W + 3.7E) = (10.6 E) + (1.8 W) . . . That adds up to 8.8 East ! That's where she ends up. That's her displacement ... 8.8 km East of where she started. Since we're only talking about displacement, we don't care HOW she got there. She might have been swimming big 20-km circles all day. We don't know. All we know is that she ended up 8.8 km East of where she started.

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3 years ago

The position-time equation for a certain train is

astraxan [27]

Answer:

$a=4.8m/s^2$

Explanation:

Hello,

In this case, since the acceleration in terms of position is defined as its second derivative:

$a=\frac{d^2x(t)}{dt^2}=\frac{d^2}{dt^2}(2.9+8.8t+2.4t^2)$

The purpose here is derive x(t) twice as follows:

$a=\frac{d^2x(t)}{dt^2}=\frac{d}{dt}(8.8+2*2.4*t)\\ \\a=4.8m/s^2$

Thus, the acceleration turns out 4.8 meters per squared seconds.

Best regards.

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4 years ago

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)