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yKpoI14uk [10]
3 years ago
15

If there was a decrease in consumer savings due to an increase in consumer spending with no increase in the money released into

circulation what will happen to the aggregate demand curve
Physics
1 answer:
lana66690 [7]3 years ago
6 0
The aggregate demand curve will also decrease. If supply is not high and there is no circulating income or monetary value that's happening in a particular market, then the demand of consumers will also go down. This is because the need for production is no longer necessary because there will be no consumers to purchase goods and services from the market.
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A 10.0 kg weather rocket generates a thrust of 230 NN . The rocket, pointing upward, is clamped to the top of a vertical spring.
blondinia [14]

Answer: 0.2m

Explanation: Firstly only the Rocket's Weight Compress the spring which can be found by

F_r=M_r*g\\F_r=10*9.81\\F_r=98.1N

According to Hooks Law

F_r=k*x\\x=F_r/k\\x=98.1/480\\x=0.2m

The part b and c of this question is done in the attachment

7 0
3 years ago
A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.10 m/s. How
Vlad [161]

Answer:

769,048.28Joules

Explanation:

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.10 m/s. How much energy was lost to air friction during this bump

The energy lost due to friction is expressed using the formula;

Energy lost  = Potential Energy + Kinetic Energy

Energy lost  = mgh + 1/2mv²

m is the mass

g is the acceleration due to gravity

h is the height

v is the speed

Substitute the given values into the formula;

Energy lost  = 56(9.8)(1400) + 1/2(56)(5.10)²

Energy lost  = 768,320 + 728.28

Energy lost  = 769,048.28Joules

<em>Hence the amount of energy that was lost to air friction during this jump is 769,048.28Joules</em>

6 0
2 years ago
I need help with this question ASAP I am on a timer
Tom [10]
First box and third box !
4 0
2 years ago
We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu
liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}

\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}

f = 30 cm

using lens formula

\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})

R_1 = R\ and\ R_2 = -R

\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})

R = (n -1)\ f

R = 2(1.5 -1)\ 30

R = 30 cm

hence, the radii of curvature is 30 cm.

3 0
3 years ago
In a hydraulic lift, the radius of the small and large pistons are
Gennadij [26K]

Answer:

the answer is 52.15cm×35.5

8 0
2 years ago
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