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3 years ago

A 5.0 kg rock is dropped from the top of a building. The speed of the rock after it has fallen for 2.2 seconds is?

Physics

1 answer:

PSYCHO15rus [73]3 years ago

5 0

5x2.2/3 Because that's how you calculate velocity.

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If cooling occurred at the bottom of a lake instead of its surface, the lake

AlladinOne [14]

No because the top has more oxygen to it which will break the particles up and have them freeze easier.

so if something froze at the bottom of the lake it would float to the top and that's why water always freezes from the top down :))

5 0

3 years ago

Read 2 more answers

30 points plz help ill do anything... literally anything.

ruslelena [56]

Answer:

1. 2.5s

Explanation:

1. For time, divide Distance / speed

25m / 10

=2.5s

3 0

3 years ago

A 21.3 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.7 cm from

Bezzdna [24]

Answer:

The magnetic field strength due to current flowing in the wire is9.322 x 10⁻⁶ T.

Explanation:

Given;

electric current, I = 21.3 A

distance of the magnetic field from the wire, R = 45.7 cm = 0.457 m

The strength of the resulting magnetic field at the given distance is calculated as;

$B = \frac{\mu_o I}{2\pi R}$

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A

$B = \frac{\mu_o I}{2\pi R}\\\\B = \frac{4\pi*10^{-7} *21.3}{2\pi(0.457)}\\\\B = 9.322 *10^{-6} \ T$

Therefore, the magnetic field strength due to current flowing in the wire is 9.322 x 10⁻⁶ T.

7 0

3 years ago

You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro

ehidna [41]

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

$F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N$

The spring obeys Hook's law:

$F=k\Delta x$

where k is the spring constant and $\Delta x$ is the stretching of the spring. Since we know $\Delta x=2.92 cm=0.0292 m$ , we can re-arrange the equation to find the spring constant:

$k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m$

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

$F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N$

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

$\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm$

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance $\Delta x$ is equal to the elastic potential energy stored by the spring, given by:

$W=U=\frac{1}{2}k\Delta x^2$