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3 years ago

(1) Predator-prey is the most important type of ecological relationship.

Physics

1 answer:

DIA [1.3K]3 years ago

4 0

1) True 2) True 3)True I hope I helped

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Why cooking pans don't have have metal handles?

Harrizon [31]

metals are conductors, meaning that heat travels through it better than insulators such as plastic (commonly used for pan handles)

Therefore, to avoid the heat burning hands, most pans don't have metal handles.

4 0

3 years ago

Read 2 more answers

What is the average velocity of a car that travels 50 kilometers due west in 0.50 hour?

Pie

Average Velocity= displacement/time Av=50/0.50 Av=100

4 0

3 years ago

The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

3 0

3 years ago

Read 2 more answers

While buying a hot plate you notice the resistance of the hot

marissa [1.9K]

Answer:

Current = 5.45amps

Power = 654 watts

Explanation:

E =120V

I =?

R = 22.02 Ohms

I= E/R

I= 120/22.02

I = 5.45AmPs

P = ? P= E x R

E = 120V P= 120x5.45

I = 5.45 AmPs P= 654W

7 0

2 years ago

A flea jumps straight up to a maximum height of 0.550 m . what is its initial velocity v0 as it leaves the ground?

Alexxx [7]

Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])

Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in

0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)

0 = [Vsub0x]^2 -10.78

10.78 = [Vsub0x]^2

Sqrt(10.78) = 3.28 m/s

3 0

3 years ago