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mihalych1998 [28]
2 years ago
12

Please helpwhy does 45° produce a max. range?​

Physics
1 answer:
lesya692 [45]2 years ago
8 0

Answer:

i do not know

Explanation:

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If a boat and its riders have a mass of 1100 kg and the boat drifts in at 1.3 m/s how much work does Sam do to stop it
andriy [413]

Answer:

-929.5Joules

Explanation:

To get the work done by sam, we will calculate the kinetic energy of sam expressed as;

KE = 1/2mv²

m is the mass = 1100kg

v is the velocity = 1.3m/s

KE = 1/2(1100)(1.3)²

KE = 550(1.69)

KE = 929.5Joules

Since Sam is opposing the direction of movement, work done by him will be a negative work i.e -929.5Joules

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3 years ago
The graph shows the solubility of several different compounds in water. According to the graph, which compound can form a satura
kumpel [21]
Answer is A. NaCIO3.
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What is the formula for amplitude?
Vesna [10]

Explanation:

In periodic motion, amplitude is half the distance between the minimum and the maximum.

A = (max - min) / 2

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3 years ago
Derek owns a farm that produces soybeans, avocados, and pecans. What is the word used to describe these items?
Novosadov [1.4K]

Answer:

crops

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dairy are milk products and those aren’t

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3 years ago
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A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi
Mamont248 [21]

Answer:

a=368.97\ m/s^2

Explanation:

Given that,

Initial angular velocity, \omega=0

Acceleration of the wheel, \alpha =7\ rad/s^2

Rotation, \theta=14\ rotation=14\times 2\pi =87.96\ rad

Let t is the time. Using second equation of kinematics can be calculated using time.

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s

Let \omega_f is the final angular velocity and a is the radial component of acceleration.

\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s

Radial component of acceleration,

a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2

So, the required acceleration on the edge of the wheel is 368.97\ m/s^2.

6 0
3 years ago
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