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bazaltina [42]
2 years ago
15

A 20-cm-long stick of m = 0.600 kg is lifted by a rope tied 5.0 cm from the upper end. The other end touches a smooth floor. The

stick makes an angle θ with the floor. Find the magnitude of the normal reaction from the floor on the stick.
Physics
1 answer:
Liula [17]2 years ago
3 0

Answer:

https://www.chegg.com/homework-help/questions-and-answers/part-20-cm-long-stick-m-0400-kg-lifted-rope-tied-70-cm-upper-end-end-touches-smooth-floor--q81268572

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Hello, how to do qn 5?:)l
Hitman42 [59]

Answer:

A

Explanation:

Please see the attached picture for the full solution.

Since we are only concerned about the decrease in gravitational potential energy of the car, we look at the decrease in height of the car as it moves from point X to point Y, instead of the distance travelled by the car.

7 0
3 years ago
Squid and Octolules are propelled by expelting water, which is maintained in a body cavity. A squid of 6.5 kg (including water i
Elina [12.6K]

Answer:

see attachment

Explanation:

4 0
3 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}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
Read 2 more answers
An object is placed at zero zero on a Number line. It moves three units to the right, then four units to the left, and then 60 u
STatiana [176]

Answer:

the displacement of the object is 5 units

Explanation:

The computation of the displacement of the object is shown below:

= Move to the right + move to the right - move to the left

= 6 units + 3 units - 4 units

= 9 units - 4 units

= 5 units

Hence, the displacement of the object is 5 units

7 0
3 years ago
Can you please give me the answers ?
rodikova [14]

Answer:

237

Explanation:

7 0
3 years ago
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