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3 years ago

Please help me. Thank you!

1 answer:

muminat3 years ago

5 0

The faster/slower one i think is faster because the colder the air, the closer the particals are, and for #2 it's because the door is condensed, and the atoms are closer, therefore meaning that the sound particals don't spread, like they would in air, they're more compact (meaning they're) easier to hear, or you just have a good listening ear.

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Please help wil give brainiest & 40p.

dexar [7]

Answer:

This is the answer

Explanation:

You can find the ans in the photo I attached.

3 0

3 years ago

A 17 kg box sitting on a shelf has a potential energy of 350 J. How high is the shelf? Round your answer to the nearest whole nu

Free_Kalibri [48]

Potential energy is mass * gravity * height. (m*g*h).

350 = 17*9.8*h <--350 is its energy, 17kg is its mass, and 9.8 is gravity's acceleration on the object. We now just need to solve for h.

h = 350/(17 * 9.8) = 2.1 meters, which, when rounded to the nearest whole meter, is 2 meters.

The shelf is 2 meters high.

3 0

3 years ago

Read 2 more answers

At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

For proton:

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

For electron:

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

For a proton we have:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

For an electron we have:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

6 0

4 years ago

A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the

Ostrovityanka [42]

Answer:

The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

Explanation:

Given;

number of turns of the flat circular loop, N = 18 turns

radius of the loop, R = 15.0 cm = 0.15 m

current through the wire, I = 0.51 A

The magnetic field through the center of the loop is given by;

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

Where;

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

$B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T$

Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

6 0

4 years ago

A worker is pushing a crate of tools up a ramp into a truck. The crate has a mass of 120 kg and is accelerating at a rate of 1.0

jolli1 [7]

Answer:

B) 0.493

Explanation:

First Crate

Formula: F=ma

120kg*1.05m/s^2= 126N

Second Crate

Formula: F=ma

3*120kg=360kg

360kg*0.71m/s^2=255.6

Find the ratio by dividing

120kg/255.6kg=0.4929

Round 0.4929=0.493

Hope this Helps!!

8 0

3 years ago