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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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Helppp plss

shutvik [7]

Answer:

a) P = 149140[w]; b) 1491400[J]; c) v = 63.06[m/s]

Explanation:

As the solution to the problem indicates, we must convert the power unit from horsepower to kilowatts.

P = 200 [hp]

$200[hp] * 745.7 [\frac{watt}{1 hp}]\\149140[watt]$

Now the power definition is known as the amount of work done in a given time

P = w / t

where:

w = work [J]

t = time [s]

We have the time, and the power therefore we can calculate the work done.

w = P * t

w = 149140 * 10 = 1491400 [J]

And finally, we can calculate the velocity using, the expression for kinetic energy

$E_{k}=w=0.5*m*v^{2}\\ where:\\v = velocity[m/s]\\m=mass=750[kg]\\w=work=1491400[J]\\$

The key to solving this problem is to recognize that work equals kinetic energy

$v=\sqrt{\frac{w}{0.5*m}} \\v=\sqrt{\frac{1491400}{0.5*750}} \\v=63.06[m/s]$

3 0

3 years ago

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

8 0

4 years ago

Help me please ()- :))

ratelena [41]

Answer:

A pardon

Explanation:

Hope this helps

8 0

3 years ago

Read 2 more answers

Find a unit vector in the direction in which f increases most rapidly at P and give the rate of chance of f in that direction; f

Burka [1]

Answer:

Check attachment for complete question

Question

Find a unit vector in the direction in which

f increases most rapidly at P and give the rate of change of f

in that direction; Find a unit vector in the direction in which f

decreases most rapidly at P and give the rate of change of f in

that direction.

f (x, y, z) = x²z e^y + xz²; P(1, ln 2, 2).

Explanation:

The function, z = f(x, y,z), increases most rapidly at (a, b,c) in the

direction of the gradient and decreases

most rapidly in the opposite direction

Given that

F=x²ze^y+xz² at P(1, In2, 2)

1. F increases most rapidly in the positive direction of ∇f

∇f= df/dx i + df/dy j +df/dz k

∇f=(2xze^y+z²)i + (x²ze^y) j + (x²e^y + 2xz)k

At the point P(1, In2, 2)

Then,

∇f= (2×1×2×e^In2+2²)i +(1²×2×e^In2)j +(1²e^In2+2×1×2)

∇f=12i + 4j + 6k

Then, unit vector

V= ∇f/|∇f|

Then, |∇f|= √ 12²+4²+6²

|∇f|= 14

Then,

Unit vector

V=(12i+4j+6k)/14

V=6/7 i + 2/7 j + 3/7 k

This is the increasing unit vector

The rate of change of f at point P is.

|∇f|= √ 12²+4²+6²

|∇f|= 14

2. F increases most rapidly in the positive direction of -∇f

∇f=- (df/dx i + df/dy j +df/dz k)

∇f=-(2xze^y+z²)i - (x²ze^y) j - (x²e^y + 2xz)k

At the point P(1, In2, 2)

Then,

∇f= -(2×1×2×e^In2+2²)i -(1²×2×e^In2)j -(1²e^In2+2×1×2)

∇f=-12i -4j - 6k

Then, unit vector

V= -∇f/|∇f|

Then, |∇f|= √ 12²+4²+6²

|∇f|= 14

Then,

Unit vector

V=-(12i+4j+6k)/14

V= - 6/7 i - 2/7 j - 3/7 k

This is the increasing unit vector

The rate of change of f at point P is.

|∇f|= √ 12²+4²+6²

|∇f|= 14

6 0

4 years ago

Read 2 more answers

How do the prefixes micro,<br> nano and pico relate to each<br> other?

In-s [12.5K]

Answer:

because they are same and their properties

8 0

3 years ago