1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
grigory [225]
3 years ago
11

Ethan drove a 1500 kg car around a circular track with a radius of 75 m. If the magnitude of the centripetal acceleration is 2 m

/s2, what is the car's speed?
A)2.25 m/s
B)20 m/s
C)37.5 m/s
D)150 m/s ​
Physics
1 answer:
Aloiza [94]3 years ago
5 0

Answer:

12.25 m/s

Explanation:

The following data were obtained from the question:

Mass (m) = 1500 Kg

Radius (r) = 75 m

Centripetal acceleration (a) = 2 m/s²

Velocity (v) =?

Centripetal acceleration (a) is related to the velocity (v) and radius (r) according to the following formula:

a = v²/r

Thus, we can obtained the velocity of the car by using the above formula as illustrated below:

Radius (r) = 75 m

Centripetal acceleration (a) = 2 m/s²

Velocity (v) =?

a = v²/r

2 = v²/75

Cross multiply

v² = 2 × 75

v² = 150

Take the square root of both side.

v = √150

v = 12.25 m/s

Thus, the speed of the car is 12.25 m/s.

You might be interested in
In the reaction 2Ca + O2 → 2CaO for every 2 Ca you will need how much O2?
Serjik [45]

Answer:

1

Explanation: that is the ratio

5 0
3 years ago
1: In a longitudinal wave, the particle displacement is __________ the direction of wave movement
jenyasd209 [6]
The answer is parallel

5 0
3 years ago
Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy
Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

P_{1}V_{1}=nRT

V_{1}=\dfrac{nRT}{P_{1}}

Put the value into the formula

V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}

V_{1}=2.62\times10^{-3}\ m^3

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}

Put the value into the formula

T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}

T_{2}=233.7\ K

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

P_{1}V_{1}=P_{2}V_{2}

V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}

Put the value into the formula

V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}

V_{2}=0.00786\ m^3

V_{2}=7.86\times10^{-3}\ m^3

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}

V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}

V_{2}=0.0057\ m^3

V_{2}=5.7\times10^{-3}\ m^3

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

6 0
3 years ago
How many kindergarteners do you think you can take on a fight before getting tired or over powered?
yulyashka [42]
Is this a serious question ?
4 0
3 years ago
Read 2 more answers
a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cl
bekas [8.4K]

Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m

Suppose, determine the time taken by the projectile to hit the ground.

We need to calculate the time

Using second equation of motion

s=ut+\dfrac{1}{2}gt^2

Where, s = vertical height of cliff

u = initial vertical velocity

g = acceleration due to gravity

Put the value in the equation

230=0+\dfrac{1}{2}\times9.8\times t^2

t=\sqrt{\dfrac{230\times2}{9.8}}

t=6.85 sec

Hence, The time taken by the projectile to hit the ground is 6.85 sec.

7 0
3 years ago
Other questions:
  • A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif
    9·1 answer
  • Assume the population of a large city like New York City is about 4 × 10^6 people, with about 1.97 people per household. Approxi
    13·1 answer
  • Three objects -- two of mass m and one of mass m -- are located at three corners of a square of edge length ℓ. find the gravitat
    12·1 answer
  • From the concepts you have learned in this module, how are you going to assess
    12·1 answer
  • Define the law of conservation of energy
    11·2 answers
  • Peregrine falcons are the fastest birds in the world, reaching max- imum speeds of up to 320 km/h (88 m/s) during a hunting stoo
    12·1 answer
  • What does attributed mean
    6·2 answers
  • A california condor is sitting on the top branch of a redwood tree and has a GPE of 12,100 J. If the condor has a mass of 8kg, w
    14·1 answer
  • GIVING BRAINLIEST PLEASE HELP!!
    15·1 answer
  • A baseball on the Moon and an identical baseball on the Earth are thrown vertically upward with the same initial velocity of 12
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!