All categories

3 years ago

A balloon and a mass are attached to a rod that is pivoted at P.

Physics

1 answer:

Brrunno [24]3 years ago

4 0

Answer:

The correct option is;

B Move both the balloon and mass 10 cm to the right

Explanation:

Given that the system is in equilibrium, we have;

Force of balloon = $F_b$ ↑

Force of mass = $F_m$ ↓

The direction of the balloon is having an upward motion which gives a clockwise moment or motion to the rod while the direction of the force of the mass weight is downwards, giving the rod an anticlockwise moment

for the rod to rotate clockwise, the moment of the balloon should be larger than that of the rod

At the present equilibrium we have;

$F_b$ × 30 = $F_m$ × 20

Therefore;

$F_m$ = 1.5× $F_b$

Moving both balloon and mass 10 cm to the right gives;

The moment of the balloon = $F_b$ × (30 - 10) = $F_b$ × 20 = 20× $F_b$ ,

The moment of the mass = $F_m$ × (20 - 10) = $F_m$ × 10

When we substitute $F_m$ = 1.5× $F_b$ in the moment equation for the mass, we have;

The moment of the mass = $F_m$ × 10 = 1.5× $F_b$ ×10 = 15× $F_b$

Therefore, the balloon now has a larger momentum than that of the mass and the rod will rotate clockwise.

You might be interested in

A fireman standing on a 15 m high ladder

ludmilkaskok [199]

Answer:

31.6 m/s

Explanation:

Mass is conserved, so the mass flow at the outlet of the pump equals the mass flow at the nozzle.

m₁ = m₂

ρQ₁ = ρQ₂

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁ πd₁²/4 = v₂ πd₂²/4

v₁ d₁² = v₂ d₂²

Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = 0 and P₂ = 0:

P₁ + ½ ρ v₁² = ½ ρ v₂² + ρgh₂

Writing v₁ in terms of v₂:

P₁ + ½ ρ (v₂ d₂²/d₁²)² = ½ ρ v₂² + ρgh₂

P₁ + ½ ρ (d₂/d₁)⁴ v₂² = ½ ρ v₂² + ρgh₂

P₁ − ρgh₂ = ½ ρ (1 − (d₂/d₁)⁴) v₂²

Plugging in values:

579,160 Pa − (1000 kg/m³)(9.8 m/s²)(15 m) = ½ (1000 kg/m³) (1 − (1.99 in / 3.28 in)⁴) v₂²

v₂ = 31.6 m/s

8 0

3 years ago

PLZ HELP ON #22-26!!!! Please explain why and how you got your answer.

AleksAgata [21]

22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2

23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.

24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s

25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m

26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.

I hope this helps!!

5 0

3 years ago

Potential energy is related to the speed of an object. True or False

adelina 88 [10]

Answer:

False

Explanation:

Potential energy is stored energy, so it is not in motion. Kinetic energy is the energy in motion.

4 0

3 years ago

Variations in the resistivity of blood can give valuable clues about changes in various properties of the blood. Suppose a medic

Elena-2011 [213]

Answer:

1.1 x 10^9 ohm metre

Explanation:

diameter = 1.5 mm

length, l = 5 cm

Potential difference, V = 9 V

current, i = 230 micro Ampere = 230 x 10^-6 A

radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m

Let the resistivity is ρ.

Area of crossection

A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2

Use Ohm's law to find the value of resistance

V = i x R

9 = 230 x 10^-6 x R

R = 39130.4 ohm

Use the formula for the resistance

$R=\rho \frac{l}{A}$

$\rho =\frac{RA}{l}$

$\rho =\frac{39130.4\times 0.05}{1.766\times 10^{-6}}$

ρ = 1.1 x 10^9 ohm metre

Explanation:

7 0

3 years ago

How can the equation of dropping and bouncing a ball have a y intercept if the ball cannot be dropped from 0 cm?

telo118 [61]

You can define the height of zero to be at any level you want.
Then the equation will give the height of the ball above that level
at any time. If you define zero height as the position of your hand
just as you drop the ball, then the equation will give negative values
for any time after the drop. If you define zero height as the floor, then
the equation will give a value of zero at the instant of every bounce,
and positive values at any other time.

For a graph of (something vs. time), the x-axis is usually time, so the
y-intercept is the height of the ball at the drop, when time begins.
That's just the height from which it's dropped ... relative to whatever
height you decide to define as zero.

7 0

3 years ago