All categories

3 years ago

A hockey puck initially travelling to the right at 34 m/s. It moves for 7 before

Physics

1 answer:

goldfiish [28.3K]3 years ago

3 0

Answer:

$x=119m$

Explanation:

Hello,

In this case, since the hockey puck was moving at 34 m/s and suddenly stopped (final velocity is zero) in 7 seconds, we can first compute the acceleration via:

$a=\frac{v_f-v_o}{t}=\frac{0m/s-34m/s}{7s}\\ \\a=-4.86m/s^2$

In such a way, we can compute the displacement via:

$x=\frac{v_f^2-v_o^2}{2a}\\ \\x=\frac{0^2-(34m/s)^2}{2*-4.86m/s^2}\\ \\x=119m$

Best regards.

You might be interested in

A person standing a certain distance from four identical loudspeakers is hearing a sound level intensity of 125 dB. What sound l

DENIUS [597]

Answer:

$\mathbf{\beta = 123.75 \ dB}$

Explanation:

From the question, using the expression:

$125 \ dB = 10 \ log (\dfrac{I}{I_o})$

where;

$I_o = 10^{-12} \ W/m^2$

$I = 10^{12.5} \times 10^{-12} \ W/m^2$

$I = 3.162 \ W/m^2$

This is a combined intensity of 4 speakers.

Thus, the intensity of 3 speakers = $\dfrac{3.162\times 3}{4}$

= 2.372 W/m²

Thus;

$\beta = 10 \ log ( \dfrac{2.372}{10^{-12}} ) \ W/m^2$

$\mathbf{\beta = 123.75 \ dB}$

7 0

3 years ago

A closely wound circular coil has a radius of 6.00 cmand carries a current of 2.65 A. How many turns must it have if the magneti

ad-work [718]

Answer:

Given:

radius of the coil, R = 6 cm = 0.06 m

current in the coil, I = 2.65 A

Magnetic field at the center, B = $6.31\times 10^{4} T$

Solution:

To find the number of turns, N, we use the given formula:

$B = \frac{\mu_{o}NI}{2R}$

Therefore,

$N = \frac{2BR}{\mu_{o}I}$

$N = \frac{2\times 6.31\times 10^{4}\times 0.06}{4\pi \times 10^{- 7}\times 2.65}$

N = 22.74 = 23 turns (approx)

8 0

3 years ago

True or false: Balanced forces can change an object's direction?

slava [35]

The statement is false. Balanced forces can NOT change the speed OR direction of an object's motion. (See Newton's #1 law of motion.)

4 0

3 years ago

Read 2 more answers

Determine tge energy in joules of a photon whose frequency is 3.55x10

JulijaS [17]

E = hf
E = 6.63×10^-34 × 3.55×10 eV
1 eV = 1.60×10^-19 J
E = 6.63×10^-34 × 3.55×10 × 1.60×10^-19
E = 3.77×10^-51 J

Hope it helped!

6 0

3 years ago

A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f

Zina [86]

To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

$A=\pi r^2$