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Lera25 [3.4K]
3 years ago
14

A car is traveling north at 17.7m/s After 12 s its velocity is 14.1m/s in the same direction. Find the magnitude and direction o

f the car's average acceleration.A2.7 m/s2, north b0.30 m/s2, south c2.7 m/s2, south d0.30 m/s2, north
Physics
1 answer:
dem82 [27]3 years ago
7 0

Answer:

Option B

0.3 m/s2 South

Explanation:

Acceleration, a=\frac {v-u}{t} where v and u are final and initial velocities respectively, t is the time taken

Substituting 14.1 m/s for v, 17.7 m/s for u and 12 s for t then

a=\frac {14.1 m/s- 17.7 m/s}{12}=-0.3 m/s^{2}

Since this is negative acceleration, it's direction is opposite hence 0.3 m/s2 South

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An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v
maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

T = \frac{PV}{nR}

T_{i} = T_{f}

\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}

Where:

P_{i}: is the initial pressure = 5.79 atm

P_{f}: is the final pressure =?

V_{i}: is the initial volume = 420 cm³

V_{f}: is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!        

3 0
3 years ago
A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o
Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let q_1 and q_2 be the charges such that

q_1+q_2=4.7

and Force between charge particles is given by

F=\frac{kq_1q_2}{r^2}

4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}

q_1\cdot q_2=0.522

put the value of q_1

q_2\left ( 4.7-q_2\right )=0.522

q_2^2-4.7q_2+0.522=0

q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}

q_2=0.114 C

thus q_1=4.586 C

3 0
3 years ago
1. Give three examples, from the lab, where potential energy was converted to kinetic energy: ​
Lubov Fominskaja [6]

Answer:

A book on a table before it falls.

A yoyo before it is released.

A raised weight.

Explanation:

These are all examples of potential energy. So I hope you can find something that is comparable from the lab.

3 0
3 years ago
A sheet of metal is 2mm wide 10cm tall and 15cm long. it was 4g. what is the density? <br>​
Hoochie [10]

Answer:

Ro = 133 [kg/m³]

Explanation:

In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

Ro = m/V

where:

m = mass [kg]

V = volume [m³]

We will convert the units of length to meters and the mass to kilograms.

L = 15 [cm] = 0.15 [m]

t = 2 [mm] = 0.002 [m]

w = 10 [cm] = 0.1 [m]

Now we can find the volume.

V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]

And the mass m = 4 [gramm] = 0.004 [kg]

Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]

3 0
2 years ago
What is frost weddging
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4 0
3 years ago
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