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4 years ago

In which parking situations should you use your parking brake?

Physics

2 answers:

vredina [299]4 years ago

6 0

Answer:

Always!!

Explanation:

I hope it help! :D

VMariaS [17]4 years ago

5 0

Answer:
-You should use your parking brake when on a hill/mountain with a huge drop because your car can’t slide off the edge.
-When there is a flood where your car is sitting because the water can’t carry your car

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If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli

Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N= $=\frac{m(V)^{2}}{R}=m*(w)^2*R$

$Friction force\\Ff = k*N\\Ff= k*m*w^2*R$

$Gravitational force \\Fg = m*g$

These must be balanced (the net force on the people will be 0) so set them equal to each other.

$Fg = Ff$

$m*g = k*m*w^2*R$

$g=k*w^{2}*R$

$w^2 =\frac{g}{k*R}$

$w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}$

There are 2*pi radians in 1 revolution so:

$RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88$

So you need about 30 RPM to keep people from falling out the bottom

7 0

4 years ago

A hoodlum throws a stone vertically downward with an initial speed v0 from the roof of a building, a height h above the ground.

Vaselesa [24]

V2=u2+2as
v2=144+600
v2=744
v=√744

6 0

3 years ago

An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate

riadik2000 [5.3K]

Answer: $6.408(10)^{-19} C$

Explanation:

This problem can be solved by the following equation:

$\Delta K=q V$

Where:

$\Delta K=7.37(10)^{-17} J$ is the change in kinetic energy

$V=115 V$ is the electric potential difference

$q$ is the electric charge

Finding $q$ :

$q=\frac{\Delta K}{V}$

$q=\frac{7.37(10)^{-17} J}{115 V}$

Finally:

$q=6.408(10)^{-19} C$

4 0

3 years ago

The blades of a fan running at low speed turn at 250 rpm. When the fan is switched to high speed, the rotation rate increases un

fgiga [73]

Answer:

1.82 rad/s².

Explanation:

Applying,

α = (ω₂-ω₁)/t..................... Equation 1

Where α = angular acceleration of the fan blades, ω₂ = final angular velocity of the fan blades, ω₁ = initial angular velocity of the fan blades, t = time.

Given: ω₂ = 350 rpm = (350×0.1047) rad/s = 36.645 rad/s. ω₁ = 250 rpm = (250×0.1047) rad/s = 26.175 rad/s, t = 5.75 s.

Substitute into equation 1

α = (36.645-26.175)/5.75

α = 10.47/5.75

α = 1.82 rad/s².

Hence the magnitude of the angular acceleration of the fan blades = 1.82 rad/s²

8 0

3 years ago

12. The diameter of a circle is 2.42m. Calculate its<br>area in proper significant figure

Sever21 [200]

Answer:

A = 4.6 [m²]

Explanation:

The area of a circle can be calculated by means of the following equation.

$A=\frac{\pi }{4} *D^{2}$

where:

A = area [m²]

D = diameter = 2.42 [m]

Now replacing:

$A=\frac{\pi }{4} *(2.42)^{2} \\A = 4.6 [m^{2} ]$

7 0

3 years ago