What is the equation of the line that is perpendicular to y= -4x +5 and passes through the point (4,-3)?

Your cat "Ms." (mass 7.00 {\rm kg}) is trying to make it to the top of a frictionless ramp 2.00 {\rm m} long and inclined upward

lbvjy [14]

Answer:

Final velocity at the top of the ramp is 6.58m/s

Explanation

Check the attachment

4 0

2 years ago

Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air

valentina_108 [34]

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: $v_f=v_0+a*t$ , where $v_f$ is the final velocity, $v_0$ is the initial velocity, $a$ the acceleration, and $t$ is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: $t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s} }{-201\frac{m}{s^2} }=1.40s$ , where $v_f=0\frac{m}{s}$ because the sled is totally stopped, $v_0=282\frac{m}{s}$ is the velocity of the sled before braking and, $a=-201\frac{m}{s^2}$ is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration: $x_f-x_0=v_0*t+\frac{1}{2}*a*t^2$ , where $x_f-x_0$ is the distance traveled, $v_0$ is the initial velocity, $t$ the time of the process and, $a$ is the acceleration of the process.

Then for this case the relationship becomes: $x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m$ .

<u>Note that the acceleration is negative because is a braking process.</u>

4 0

3 years ago

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Solve the inequality 2(n+3) – 4<6. Then graph<br> the solution.

Aloiza [94]

The solution is 22 2(n+3)-4&6

6 0

2 years ago

Suppose that you measure a galaxy's redshift, and from the redshift you determine that its recession velocity is 30,000 (3×10^4)

anyanavicka [17]

Answer:

1.4 billion light years away

Explanation:

v = Recessional velocity = 30000 km/s[/tex]

$H_0$ = Hubble constant = $\frac{65}{3.2\times 10^6}\ ly$

D = Distance to the galaxy

According to Hubble's law

$v=H_0D\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{3\times 10^{4}}{65}\times 3.2\times 10^6\\\Rightarrow D=1476923076.92307\ ly\\\Rightarrow D=1.4\times 10^9\ ly$