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dexar [7]
3 years ago
15

While traveling along a highway a driver slows from 24 m/s to 15 m/s in 12 seconds. What is the automobile’s acceleration?

Physics
1 answer:
Nimfa-mama [501]3 years ago
6 0

Average Acceleration = (change in speed) / (time for the change)

Change in speed = (ending speed) - (starting speed)

                             =   15 m/s - 24 m/s  =  -9 m/s

Acceleration =  (-9 m/s) / (12 sec) =  - 0.75 m/s² .

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Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We
Katen [24]

Answer:

q=\frac{mg}{E_o}              

Explanation:

Given:

Charge = <em>q</em>

Electric field strength =E_o

weight of the droplet = <em>mg</em>

The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.

electric force on charged droplet, F=qE_o

This is balanced by the weight, mg

Equating the two:

qE_o=mg\\\Rightarrow q=\frac{mg}{E_o}

4 0
4 years ago
Josh starts his sled at the top of a 2.9-m-high hill that has a constant slope of 25∘. After reaching the bottom, he slides acro
algol13

Answer:

S=48.29 m

Explanation:

Given that the height of the hill h = 2.9 m

Coefficient of kinetic friction between his sled and the snow μ = 0.08

Let u be the speed of the skier at the bottom of the hill.

By applying conservation of energy at the top and bottom of the inclined plane we get.

Potential Energy=kinetic Energy

mgh = (1/2) mu²

u² = 2gh

u²=2(9.81)(2.9)

   =56.89

u=7.54 m/s

a = - f / m

a = - μ*m*g / m

a = - μg

From equation of motion

v²- u² = 2 -μ g S

v=0 m/s

-(7.54)²=-2(0.06)(9.81)S

S=48.29 m

3 0
3 years ago
Why is the picture above not an accurate representation of charges and their field lines?
pochemuha
Because field lines don’t all go in to the pole. The bottom half go outwards not inwards.
7 0
3 years ago
Read 2 more answers
Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
3 years ago
4.39 moles of gas in a box has a pressure of 2.25 atm at temperature of 385K. What is the volume of the box?
bazaltina [42]

Answer:

0.0619 m^3

Explanation

number of moles = n = 4.39 mol

pressure = P = 2.25 atm =2.25×1.01×10^5 Pa= 2.27×10^5 Pa

Molar gas constant =R = 8.31 J/(mol K)

Temperature T= 385K

volume of gas = V =?

BY GENERAL GAS LAW WE HAVE

PV = nRT

or V = nRT/P

or V = (4.39×8.31×385)/(2.27×10^5)

V = 0.0618728

V =  0.0619 m^3

5 0
4 years ago
Read 2 more answers
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