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3 years ago

A 6,000N is applied to a formula one car that weighs 500kg. What is the car's acceleration?

Physics

1 answer:

Vesna [10]3 years ago

8 0

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question

f = 6000 N

m = 500 kg

We have

$a = \frac{6000}{500} = \frac{60}{5} = 12 \\$

We have the final answer as

Hope this helps you

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A police car parked on the side of the highway emits a 1200 Hz sound that bounces off a vehicle farther down the highway and ret

emmasim [6.3K]

f' = frequency observed by the police car after sound reflected from the vehicle and comes back to police car = 1250 Hz

f = frequency emitted by the police car = 1200 Hz

V = speed of sound = 340 m/s

v = speed of vehicle = ?

frequency observed by the police car is given as

f' = f (V + v)/(V - v)

inserting the values in the above equation

1250 = 1200 (340 + v)/(340 - v)

v = 6.9 m/s

4 0

3 years ago

Read 2 more answers

A hammer strikes one end of a thick iron rail of length 8.80 m. A microphone located at the opposite end of the rail detects two

stepladder [879]

Answer:

ΔT = 0.02412 s

Explanation:

We will simply calculate the time for both the waves to travel through rail distance.

FOR THE TRAVELING THROUGH RAIL:

$T_{rail} = \frac{Distance}{Speed\ of\ Sound\ in\ Rail}\\\\T_{rail} = \frac{8.8\ m}{5950\ m/s}\\\\T_{rail} = 0.00148\ s$

FOR THE WAVE TRAVELING THROUGH AIR:

$T_{air} = \frac{Distance}{Speed\ of\ Sound\ in\ Air}\\\\T_{air} = \frac{8.8\ m}{343\ m/s}\\\\T_{air} = 0.0256\ s$

The separation in time between two pulses can now be given as follows:

$\Delta T = T_{air}-T_{rail} \\\Delta T = 0.0256\ s - 0.00148\ s\\$

3 0

3 years ago

During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

geniusboy [140]

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

$a_t$ = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

$\omega_i$ = Initial angular velocity = 95 rad/s

$\theta$ = Angle of rotation

$\omega_f$ = Final angular velocity

t = Time taken

Angular acceleration is given by

$\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2$

The angular acceleration is -24.28571 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev$

The number of revolutions is 29.57239

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-24.28571}\\\Rightarrow t=3.91176\ s$

The time it takes for the car to stop is 3.91176 seconds

Linear distance

$s=r\theta\\\Rightarrow s=0.28\times 185.80885\\\Rightarrow s=52.026478\ m$

The distance the car travels is 52.026478 m

8 0

3 years ago

A motorist enters a freeway at 45 km/h and accelerates uniformly to 99 km/h. From the odometer in the car, the motorist knows th

Marat540 [252]

Answer:

a) 19440 km/h²

b) 10 sec

Explanation:

v₀ = initial velocity of the car = 45 km/h

v = final velocity achieved by the car = 99 km/h

d = distance traveled by the car while accelerating = 0.2 km

a = acceleration of the car

Using the kinematics equation

v² = v₀² + 2 a d

99² = 45² + 2 a (0.2)

a = 19440 km/h²

t = time required to reach the final velocity

Using the kinematics equation

v = v₀ + a t

99 = 45 + (19440) t

t = 0.00278 h

t = 0.00278 x 3600 sec

t = 10 sec

5 0

3 years ago

Sharon throws a 0.20 Kg with an acceleration of 10 m/s/s.

OlgaM077 [116]

Force=A×M
10m/s×0.20kg
=2Newton

4 0

3 years ago