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Paha777 [63]
3 years ago
9

11. You want to calculate the displacement of an object thrown over a bridge. Using -10 m/s2 for acceleration due to gravity, wh

at would be the total displacement of the object if it took 8 seconds before hitting the water?
Physics
1 answer:
Natalka [10]3 years ago
3 0

Displacement is d  



Vf² = Vi² + 2 g d  



(-20²) = (+10²) + 2 (-9.8) d  



-19.6 d = 300  



d = -15.3 m  



negative means lower



time is t  



d = Vi t + 1/2 g t²




-15.3 = 10 t + (-4.9) t²




4.9 t² - 10 t -15.3 = 0  



t = 3.06 s



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Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh
Alinara [238K]

Answer:

a) k_{avg}=6.22\times 10^{-21}

b) k_{avg}=8.61\times 10^{-21}

c)  k_{mol}=3.74\times 10^{3}J/mol

d)   k_{mol}=5.1\times 10^{3}J/mol

Explanation:

Average translation kinetic energy (k_{avg}) is given as

k_{avg}=\frac{3}{2}\times kT    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8  

k_{avg}=6.22\times 10^{-21}J

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416  

k_{avg}=8.61\times 10^{-21}J

c ) The translational kinetic energy per mole of an ideal gas is given as:

       k_{mol}=A_{v}\times k_{avg}

here   A_{v} = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}

          k_{mol}=3.74\times 10^{3}J/mol

d) now at T = 143° C

        k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}

          k_{mol}=5.1\times 10^{3}J/mol

8 0
3 years ago
1. A toy car is marked to go at 0.5m/s, which of the following sets of
Lostsunrise [7]

Answer:

b. 0.6m/s, 0.7m/s, 0.61m/s, 0.62m/s

Explanation:

Precision of a measurement is the closeness of the experimental values to one another. Hence, experimental measurements are said to be precise if they are close to each other irrespective of how close they are to the accepted value. Precision can be determined by finding the range of each experimental value. The measurement with the LOWEST RANGE represents the MOST PRECISE.

Note: Range is the highest value - lowest value

Set A: 1.5 - 0.8 = 0.7

Set B: 0.7 - 0.6 = 0.1

Set C: 2.4 - 2.0 = 0.4

Set D: 3.1 - 2.9 = 0.2

Set B has the lowest range (0.1), hence, represent the most precise value.

8 0
3 years ago
The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p
NISA [10]

Answer:

a)  The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be v_{s} = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

t = \frac{d}{v_{s} }

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be v_{r} = 2.5 m/s

S = v_{r} t

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}

d) Downstream velocity of the swimmer, v_{y} = v_{s} sin \theta\\

v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s

The vertical displacement is given by, y = v_{y} t

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s

The downstream horizontal distance of the swimmer, x = v_{x} t

x = 1.6 * 66.67

x = 106.67 m

7 0
3 years ago
Question 9 (28 points)
ladessa [460]
Im sure the answer is letter B
5 0
3 years ago
Read 2 more answers
A conducting loop has an area of 0.065 m2 and is positioned such that a uniform magnetic field is perpendicular to the plane of
aev [14]

Answer:

initial magnetic field  1.306 T

Explanation:

We have given area of the conducting loop A=0.065m^2

Emf induced = 1.2 volt

Initial magnetic field B = 0.3 T

Time dt = 0.087 sec

We know that induced emf is given by e=\frac{d\Phi }{dt}=-A\frac{db}{dt}

1.2=0.065\times \frac{db}{0.087}

db=1.606T

So initial magnetic field = 1.606-0.3= 1.306 T

5 0
3 years ago
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