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ohaa
2 months ago
10

In 1991 four English teenagers built an eletric car that could attain a speed of 30.0m/s. Suppose it takes 8.0s for this car to

accelerate from 18.0m/s to 30.0m/s. What is the magnitude of the car's acceleration?
Physics
2 answers:
inna [3.1K]2 months ago
8 0

Answer:

Acceleration of the vehicle, a=1.5\ m/s^2

Explanation:

The initial velocity of the vehicle is u = 18 m/s

The final velocity of the vehicle is v = 30 m/s

The time duration for this change is t = 8 s

To find the acceleration, denoted as a, we can apply the first equation of motion.

v=u+at

Where

a represents the car’s acceleration.

a=\dfrac{v-u}{t}

a=\dfrac{30-18}{8}      

a=1.5\ m/s^2

Thus, the car's acceleration is 1.5\ m/s^2. This is the solution we are looking for.                                          

Sav [3.1K]2 months ago
6 0

a=Δv/Δt=(30.0-18.0)/8.0=12.0/8.0=1.5 m/s²


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Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
kicyunya [3294]

Result:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The electromagnetic attraction between the electron and the proton in the nucleus is equivalent to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k represents the Coulomb constant

e denotes the charge of the electron

e denotes the charge of the proton in the nucleus

r signifies the distance from the electron to the nucleus

v indicates the velocity of the electron

is the mass of the electron

Rearranging for v, we determine

v=\sqrt{k\frac{e^2}{m_e r}}

Inside a hydrogen atom, the distance separating the electron from the nucleus is roughly

r=5.3\cdot 10^{-11}m

while the mass of the electron is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

By plugging in the values into the formula, we achieve

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
1 month ago
A car with an initial velocity of 16.0 meters per second east slows uniformly to 6.0 meters per second east in 4.0 seconds. What
serg [3582]
(6-16)/4.0=-2.5 m/s²
The car's acceleration is -2.5 m/s²
5 0
2 months ago
The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va
serg [3582]

Answer:

10000 V

0.00225988700565 m²

8\times 10^{-12}\ F

Explanation:

E = Electric field = 4\times 10^6\ V/m

d = Distance = 2.5 mm

Q = Charge = 80 nC

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

The potential difference is calculated as

V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V

The potential difference across the plates amounts to 10000 V

Area is determined by

A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2

The area of each plate measures 0.00225988700565 m²

Capacitance is determined by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F

The capacitance is 8\times 10^{-12}\ F

4 0
1 month ago
What is the kinetic energy of a 100 kg object that is moving with a speed of 12.5m/s
Softa [3030]

The formula for the kinetic energy of any object in motion is

                           (1/2) (mass) (velocity²).

For the object you've mentioned, it translates to

                            (1/2) (100 kg) (12.5 m/s)²

                         =      (50 kg)  (156.25 m²/s²)

                         =              7,812.5 joules  
______________________________

Beware that your attachment is heavily blurred and unreadable.

7 0
1 month ago
A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108
Ostrovityanka [3204]

Answer: The calorimeter's heat capacity is 6.72J/g^oC

Explanation:

This scenario assumes the amount of heat lost by the hot object equals the amount of heat gained by the cold object.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat capacity of water = 4.184J/g^oC

c_2 = specific heat capacity of calorimeter =?

m_1 = mass of water = 108.7 g

m_2 = mass of calorimeter = 108.7 g

T_f = final temperature of the mixture = 35.0^oC

T_1 = initial temperature of the water = 60.2^oC

T_2 = initial temperature of calorimeter = 19.3^oC

Now substituting all provided values into the formula, we obtain

(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC

c_2=6.72J/g^oC

Hence, the heat capacity of the calorimeter is 6.72J/g^oC

3 0
2 months ago
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