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4 years ago

Can someone pls help ASAP:(

Physics

1 answer:

diamong [38]4 years ago

6 0

I’m not sure but I think it’s
△ m=5 and △= -3 and so

Answer: 5/△-3 m/s

So sorry if it’s wrong

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Project managers and their teams must keep in mind the effects of any project on the interests and needs of the entire system or

amm1812

Answer:True

Explanation:

True

Even though it is easier to concentrate on the urgent and sometimes specific issues of a given project, project managers and all other personnel must bear in mind the influence of any project on the wants and needs of the entire system or organization.

A Project manager must perform following duties

Planning Project resources
Lead the team
Time management
Budget
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3 years ago

At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until

Rama09 [41]

Answer:

(A) 570 rad

(B) 10 s

Explanation:

The equations of motion for circular motions are used.

Initial angular velocity, $\omega_0 = 30.0 \text{ rad/s}$

Angular acceleration, $\alpha =35.0 \text{ rad/s}^2$

(A)

At t = 2.00 s, the angular displacement, θ, is given by

$\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}$

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

$\omega = \omega_0+\alpha t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}$

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

$\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t$

Here, $\theta_2=440$ (the angular displacement during deceleration)

The subscripts, i and f, on ω denote the initial and final angular velocities during deceleration.

$\omega_i = 100$

$\omega_f = 0$

$t = \dfrac{2\theta_2}{\omega_i} = \dfrac{2\times400}{100} = 8\ \text{s}$

This is the time for deceleration. The deceleration began at t = 2 s.

Hence, the wheel stops at t = 2 + 8 = 10 s.

(C)

The deceleration is given by

$\alpha_R = \dfrac{\omega_f-\omega_i}{t} = \dfrac{0-100}{8} = -12.5\text{ rad/s}^2$

The negative sign appears because it is a deceleration.

4 0

3 years ago

In the figure , if Q = 30 uC , q = 5.0 uC , and d = 30 cm , what is the magnitude of the electrostatic force on g?

lana [24]

Answer:

$F = k\frac{q_1q_1}{ {d}^{2} } \\ = 9.0 \times {10}^{9} (N {m}^{2} {C}^{ - 2}) \frac{30 \times {10}^{ - 6} (C)5.0 \times {10}^{ - 6} (C)}{ {0.3}^{2} } N$

3 0

3 years ago

What type of interaction allows the sidewalk to heat up and cook the egg?

Bogdan [553]

Answer:

Adsorption

Explanation:

Sidewalk cooking of egg is very popular in America. In summer people release fireworks in night sky and cook egg on the concrete sidewalk to check the level of temperature. When sunlight fall on the sidewalk most of the light is reflected back but some darker material adsorbs some photon, and when these photons are transferred to egg molecules it causes vibration among them and produce heat and cook egg.

7 0

3 years ago

Read 2 more answers

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lana [24]

Answer:

the answer is b I believe if you do the math right

5 0

3 years ago