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grandymaker [24]
2 years ago
15

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

ume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 29.0 m/s.
Physics
1 answer:
vivado [14]2 years ago
4 0

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

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Find the frequency of a wave of wavelength 2.5m and speed 400 m/s
Sedaia [141]

Answer:

The frequency of wave is 160Hz.

Explanation:

Given that the formula of speed is V = f×λ where V represents speed, f is frequency and λ is wavelength.

So first thing, you have to make frequency the subject by dividing wavelength on both sides :

v = f \times λ \:

v \div  λ =  f \times λ \div λ

f =  \frac{v}{λ}

Next you have to substitute the value of v and f into the formula :

Let λ = 2.5m,

Let v = 400m/s,

f =  \frac{400}{2.5}

f = 160Hz

7 0
2 years ago
the gas in a balloon has P=100000 pa and v=0.0279 m^3. if the pressure increases to 120000 pa at constant temperature, what is t
mash [69]

Answer:

New volume of the baloon is 0.02325m^3

Explanation:

To answer this question we need to know the ideal gas law, which says:

p•V = n•R•T

p is pressure, V is volume, n is amount of substance (in moles), R is constant value and T is temperature.

Since it's stated that n and T are constant, and we know that R is a constant too, that means that p•V = constant value. Basically, that means that p1•V1 (pressure and volume before the pressure increase) equals to p2•V2 (pressure and volume after the pressure increase).

That means that:

100000 Pa • 0.0279 m^3 = 120000 Pa • V2. Next, V2= 100000•0.0279/120000. So, V2=0.02325m^3.

6 0
3 years ago
Read 2 more answers
Technician A says that a power-assisted brake system reduces stopping distances compared to a nonpower-assisted brake system. Te
yanalaym [24]

Answer:

Technician B

Explanation:

here on analyzing both the statements from technician A and technician B. The Statement from Technician B is more logical and correct. That the power-assisted brake system reduces the force that the driver must exert on the brake pedal.

The power-assisted brake system does not reduce the distance of stopping. What it does is it reduces the force to be applied by the driver. Thus, making the drive more comfortable.

4 0
3 years ago
An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f
sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

f=\frac{v_s}{2L}         (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

f=\frac{343m/s}{2(0.47228m)}=362.36Hz      

Closed tube:

f'=\frac{v_s}{4L'}

L': length of the closed tube = 0.702821m

f'=\frac{343m/s}{4(0.702821m)}=122.00Hz

Next, you use the following formula for the beat frequency:

f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0
3 years ago
A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00
vivado [14]

Answer:

7.5 m

Explanation:

v = initial speed of the ball = 8 m/s

\theta = angle of launch = 40° deg

Consider the motion along the vertical direction :

v_{oy} = initial velocity along vertical direction = v Sin\theta = 8 Sin40 = 5.14 m/s

a_{y} = acceleration along vertical direction = - 9.8 m/s²

t = time of travel

y = vertical displacement = - 1 m

Using the kinematics equation

y = v_{oy}t + (0.5)a_{y}t^{2}

- 1 = (5.14)t + (0.5)(- 9.8)t^{2

t = 1.22 sec

Consider the motion along the horizontal direction :

v_{ox} = initial velocity along horizontal direction = v Sin\theta = 8 Cos40 = 6.13 m/s

a_{x} = acceleration along vertical direction = 0 m/s²

t = time of travel = 1.22 sec

x = horizontal displacement = ?

Using the kinematics equation

x = v_{ox}t + (0.5)a_{x}t^{2}

x = (6.13)(1.22) + (0.5)(0)(1.22)^{2

x = 7.5 m

4 0
3 years ago
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