What happens to the air pressure inside a balloon when the balloon is squeezed to half its volume at constant temperature?

5. Why does a Ntion try to influence the value of its currency?

Mazyrski [523]

Answer:

If inflation is relatively lower than competitors, then the countries goods will become more attractive and demand will rise. Lower inflation tends to increase the value of the currency in the long term.

Explanation:

7 0

2 years ago

In the woman's mouth if

Tpy6a [65]

Answer:

hi mate,

interesting question, first of all the pressure is determined by using the following formula:

Pg = p * G * h

where p is the density of the liquid, G is the gravity and h is the height difference, in you case you have:

p = 1015 kg/m3

G = 9.8m/s2

h = 0.085 m

insert these values into the equation above:

Pg = 1015 kg/m3 * 9.8m/s2 * 0.085 m = 849.81 kg·m-1·s-2 or 849.81 pascal

hope it helps, :-)

please mark me as brainliest

7 0

3 years ago

In a car, 75 percent of the chemical energy of gasoline is lost as thermal

LenaWriter [7]

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3 0

4 years ago

The magnetic flux through each turn of a 110-turn coil is given by ΦB = 9.75 ✕ 10−3 sin(ωt), where ω is the angular speed of the

Xelga [282]

Answer:

Explanation:

Given that a coil has a turns of

N = 110 turns

And the flux is given as function of t

ΦB = 9.75 ✕ 10^-3 sin(ωt),

Given that, at an instant the angular velocity is 8.70 ✕ 10² rev/min

ω = 8.70 ✕ 10² rev/min

Converting this to rad/sec

1 rev = 2πrad

Then,

ω = 8.7 × 10² × 2π / 60

ω = 91.11 rad/s

Now, we want to find the induced EMF as a function of time

EMF is given as

ε = —NdΦB/dt

ΦB = 9.75 ✕ 10^-3 sin(ωt),

dΦB/dt = 9.75 × 10^-3•ω Cos(ωt)

So,

ε = —NdΦB/dt

ε = —110 × 9.75 × 10^-3•ω Cos(ωt)

Since ω = 91.11 rad/s

ε = —110 × 9.75 × 10^-3 ×91.11 Cos(91.11t)

ε = —97.71 Cos(91.11t)

The EMF as a function of time is

ε = —97.71 Cos(91.11t)

Extra

The maximum EMF will be when Cos(91.11t) = -1

Then, maximum emf = 97.71V

8 0

3 years ago

A right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 lb/ft3

tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ $ft^{3}$ to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = $\pi r^{2}$ = $\pi 4^{2}$ =16 $\pi$ $ft^{2}$

Now,

Work done required = $\int_{0}^{8} 52\times 16\pi (21 - y)dy$

= $832\pi \int_{0}^{8} (21 - y)dy$

= 832 $\pi($ $[ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\$ )

= 113152 $\pi$ = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0

4 years ago