You can use one or more of your five senses to make ______ during an inquiry activity

1 answer:

7 0

<span>You can use one or more of your five senses to make observations during an inquiry activity.
we have nervous system which has special sensory system, we have sensory organs and senses, the five senses are touch, smell, hearing, sight and taste. every sense is related to an organ, like we can touch with hands, hear with ears, sight is related to eyes, we can smell with nose and taste with tongue.

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A car with a mass of 2000 kg is moving around a circular curve at a uniform velocity of 25 m/s per second. The curve has a radiu

IrinaK [193]

The centripetal force is:
F = mv² / R
Where:
m: mass of the object
v: object speed
R: radius of the curve.
We have to:
m = 2000kg
v = 25 m / s
R = 80 meters.
Then the centripetal force acting on the vehicle is:
F = (2000kg * (25m / s) ²) / 80m
F = 15625 N

4 0

3 years ago

A loaded barge has a mass of 1 500 000 kg and is traveling at 3 m/s. If a tugboat applies an opposing force of 12 000 N for 10 s

yan [13]

Answer:

Explanation:

Initial momentum is 1.5e6(3) = 4.5e6 kg•m/s

An impulse results in a change of momentum

The tug applied impulse is 12000(10) = 120000 N•s or 0.12e6 kg•m/s

The remaining momentum is 4.5e6 - 0.12e6 = 4.38e6 kg•m/s

The barge velocity is now 4.38e6 / 1.5e6 = 2.92 m/s

The tug applies 0.012e6 N•s of impulse each second.

The initial barge momentum will be zero in

t = 4.5e6 / 0.012e6 = 375 s or 6 minutes and 15 seconds

To stop the barge in one minute(60 s), the tug would have to apply

4.5e6 / 60 = 75000 N•s /s or 75 000 N

5 0

2 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .