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Bond [772]
3 years ago
6

1) Sketch (or scan or photocopy if you prefer) the plot below twice on your homework sheet. a) On one plot, draw the sum of the

two vectors, A+B, and on the second, draw the difference, A-B. b) Calculate the magnitudes of both, in units of the grid spacing.
Physics
1 answer:
miskamm [114]3 years ago
5 0

Answer:

yes

Explanation:

yes

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a car with a mass of 2,000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. the curve
Ber [7]
Fc=mv^2/r so we get 

2000kg*(25m/s)^2/(80m)= 15625N of force 

hope this helps! Thank You!!

4 0
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Segmented mirrors sag under their own weight. their optical shape must be controlled by computer-driven thrusters under the mirr
madam [21]
Active Optics.


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6 0
3 years ago
A mass of gas has a volume of 4m3, a temperature of 290k, and an absolute pressure of 475 kpa. When the gas is allowed to expand
motikmotik

Given:

V1 = 4m3

T1 = 290k

P1 = 475 kpa = 475000 Pa

V2 = 6.5m3

T2 = 277K

Required:

P

Solution:

n = PV/RT

n = (475000 Pa)(4m3) / (8.314 Pa-m3/mol-K)(290k)

n = 788 moles

P = nRT/V

P  = (788 moles)(8.314 Pa-m3/mol-K)(277K)/(6.5m3)

P = 279,204 Pa or 279 kPa

4 0
3 years ago
Read 2 more answers
Air Pressure ______ With Elevation. (A) Increases (B) Decreases (C) Does Not Change
Mnenie [13.5K]
Air pressure ''Decreases'' with Elevation. So B) would be your answer.
8 0
3 years ago
Read 2 more answers
A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.
Airida [17]

Answer:

a) v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} ), b) \Delta K = 9.559\times 10^{-5}\,J

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v

The final velocity is:

v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )

b) The change in the kinetic energy of the 13.5 g coin is:

\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]

\Delta K = 9.559\times 10^{-5}\,J

4 0
3 years ago
Read 2 more answers
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