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vivado [14]
2 years ago
11

5. In a local bar, a customer slides an empty beer mug on the counter for a refill. The bartender does not see the mug, which sl

ides off the counter and strikes the floor 1.4 m from the base of the counter. If the height of the counter is 0.86 m,
(a) with what speed did the mug leave the counter
(b) what was the direction of the mug's velocity just before it hit the floor?​
Physics
1 answer:
pishuonlain [190]2 years ago
5 0

Answer:

sorry about this question

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A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of
balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg

∴ f=\mu mg=0.490\times 90\times 9.8=432.18\ N

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

W=-fd=-432.18d

Now, change in kinetic energy is given as:

\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J

Therefore, from work-energy theorem,

W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m

Hence, the skater covers a distance of 15 m before stopping.

7 0
3 years ago
A normal walking speed is around 2.0 m/s . how much time t does it take the box to reach this speed if it has the acceleration 5
creativ13 [48]

Given:

u(initial velocity)=0

a=5.54m/s^2

v(final velocity)=2 m/s

v=u +at

Where v is the final velocity.

u is the initial velocity

a is the acceleration.

t is the time

2=0+5.54t

t=2/5.54

t=0.36 sec


6 0
3 years ago
A shopping cart rolls from x=19.9m to x=5.40m with an average velocity of -0.418m/s. How much time did it take? (unit=s)
andrew11 [14]

Given

initial position = Xi= 19.9m

Final position Xf = 5.4m

Average velocity= Va = -0.418m/s

it shows displacement is reverse.

To find  t=?

As   Va = (Xf- Xi) / t

t = (Xf-Xi) / ( Va)

t = ( 5.4-19.9) / (-0.418)

t = (-14.5 ) / (-0.418)   (-ve sign cancel out at numerator and denominator)

t =34.69 s

8 0
3 years ago
How much does the speed of a car increase if it accelerates
Nata [24]

Answer:

12.5 m/s

Explanation:

a = Δv / Δt

2.5 m/s² = Δv / 5 s

Δv = 12.5 m/s

4 0
3 years ago
What tension would you need to make a middle c (261.6 hz) fundamental mode on a 1 m string (for example, on a harp)? the linear
Allisa [31]
The frequency of middle C on a string is
f = 261.6 Hz.

The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
   = 0.002 kg/m

The length of the string is L = 1 m.

Let T =  the tension in the string (N).
The velocity of the standing wave is
v= \sqrt{ \frac{T}{\rho} }

In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
\sqrt{ \frac{T}{\rho} } =f \lambda = fL \\\\ \frac{T}{\rho} = (fL)^{2} \\\\ T = \rho (fL)^{2}

From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
   = 136.87 N

Answer: 136.9 N (nearest tenth)

4 0
3 years ago
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