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umka2103 [35]
3 years ago
11

A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski

dding. a) what is the normal force on the tires exerted by the pavement. b) what is the frictional force exerted by the pavement on the tires?
Physics
1 answer:
Crazy boy [7]3 years ago
6 0

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

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olganol [36]

Answer:

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Explanation:

8 0
2 years ago
A helium-filled balloon is launched when the temperature at ground level is 27.8°c and the barometer reads 752 mmhg. if the ball
ololo11 [35]
The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
p = pressure
V = volume
T = temperature.

Note that
7.5006 x 10⁻³ mm Hg = 1 Pa
1 L = 10⁻³ m³

Given:
At ground level,
p₁ = 752 mm Hg
     = (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
     = 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
     = 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
     = 300.8 K

At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
     = 9.7326 x 10³ Pa
T₂ = 235 K

If the volume at  36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
     = (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
     = 762.15 m³

Answer: 762.2 m³  
3 0
3 years ago
A 150kg person stands on a compression spring with spring constant 10, 000 N/m and norminal.Length of 0.50.What is the togal len
Eva8 [605]

Answer:

Total length of spring 0.647 m

Explanation:

We have given mass of the person m = 150 kg

Acceleration due to gravity g=9.8m/sec^2

Spring constant k = 10000 N/m

Nominal length of spring = 0.50

According to hook's law

mg=kx

150\times 9.8=10000\times x

x = 0.147 m

So total length of spring = 0.50+0.147 = 0.647 m

4 0
3 years ago
If an airplane traveled 1800 km in 2 hours<br> what is its average speed
Vesnalui [34]

Answer:

900km h

Explanation:

duh

8 0
2 years ago
An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m
Aneli [31]

Answer:

\displaystyle y_m=3.65m

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of v_o and \theta\\ as the initial speed and angle, then we have

\displaystyle v_x=v_o\ cos\theta

\displaystyle v_y=v_o\ sin\theta-gt

\displaystyle x=v_o\ cos\theta\ t

\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}

If we want to know the maximum height reached by the object, we find the value of t when v_y becomes zero, because the object stops going up and starts going down

\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt

Solving for t

\displaystyle t=\frac{v_o\ sin\theta }{g}

Then we replace that value into y, to find the maximum height

\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2

Operating and simplifying

\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}

We have

\displaystyle v_o=20\ m/s,\ \theta=25^o

The maximum height is

\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}

\displaystyle y_m=3.65m

7 0
3 years ago
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