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frosja888
2 months ago
11

A boat's capacity plate gives the maximum weight and/or number of people the boat can carry safely in certain weather conditions

. what are these conditions?
Physics
1 answer:
Maru [3.3K]2 months ago
5 0
• wind
• snow
• tides (both high and low)
• thunderstorms, including lightning storms
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What is the Physics Primer?
kicyunya [3294]

Answer:

A. a collection of mathematical topics that are pertinent to basic physics.

Explanation:

The physics primer is not the same as the comprehensive online mathematics textbooks. Instead, it comprises topics in mathematics that challenge students and are noteworthy.

Therefore, it can be understood as the framework for resolving physics-related problems. Thus, mathematical skills are integral within physics courses, serving as a preparatory tool for success.

In summary, it represents a compilation of mathematical subjects that are relevant to foundational physics.

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2 months ago
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Your school science club has devised a special event for homecoming. You've attached a rocket to the rear of a small car that ha
inna [3103]

La force agissant pendant 9 s et la décélération pendant 12 - 9 = 3 s.

Distance totale parcourue = 990 m

vitesse initiale u = 0

Distance parcourue pendant l'accélération

s₁ = 1/2 a 9² où a est l'accélération

= 40.5 a

vitesse finale après 9 s

v = at = 9a

pendant la décélération

v² = u² - 5 x s₂

0 = (9a)² - 5 s₂

s₂ = 16.2 a²

Distance parcourue pendant la décélération = 16.2 a²

s₁ + s₂ = 990

40.5 a + 16.2 a² = 990

16.2 a² + 40.5 a - 990 = 0

a = 6.5

4 0
1 month ago
A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Sav [3153]

Response:

A=0.199

Clarification:

We know that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=

\nu=1.2Hz

Energy for oscillation is 0.51 J

To determine the amplitude of oscillations.

Energy for oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Therefore, the amplitude of oscillation=A=0.199

4 0
1 month ago
Add a third force that will cause the object to remain at rest. Label the new force F⃗ 3. Draw the vector starting at the black
kicyunya [3294]
The new force F3 is added in the same direction as F2. To analyze the forces acting on an object in this scenario, we observe that they operate along the vertical axis, with F1 acting upward and F2 downward. To determine the necessary vector F3 to counteract the net force, it's important to calculate the length difference between F1 and F2. The direction of F3 will match that of the smaller force. If F2 is less than F1, F3 will align with F2.
4 0
23 days ago
During a snowball fight two balls with masses of 0.4 and 0.6 kg, respectively, are thrown in such a manner that they meet head-o
Yuliya22 [3333]

Answer:

The snowball's speed after the impact is 3 m/s

Explanation:

Given the following:

mass of each ball

m₁ = 0.4 Kg

m₂ = 0.6 Kg

initial speed of both balls = v₁ = 15 m/s

Speed of 1 Kg mass post-collision =?

Applying conservation of momentum

m₁ v₁ - m₂ v₁ = (m₁+m₂) V

A negative velocity indicates that the second ball moves in the opposite direction.

0.4 x 15 - 0.6 x 15 = (1) V

Therefore,

V = - 3 m/s

Consequently,

The snowball's speed following the collision is 3 m/s

8 0
1 month ago
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