Ask question

Ask question

All categories

2 years ago

12

A race car has a centripetal acceleration of 15.625 m/s2 as it goes around a curve. If the curve is a circle with radius 40 m, w

hat is the speed of the car?

1 answer:

myrzilka [38]2 years ago

8 0

The centripetal acceleration is given by
$a_c = \frac{v^2}{r}$
where v is the tangential speed and r the radius of the circular orbit.

For the car in this problem, $a_c = 15.625 m/s^2$ and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
$v= \sqrt{a_c r}= \sqrt{(15.625 m/s^2)(40 m)}=25 m/s$

You might be interested in

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

2 years ago

I need help!!!!<br> Please and thank u!

alexandr1967 [171]

Answer:

False

True

False

True

False

True

Explanation:

5 0

3 years ago

When joshua brakes his speeding bicycle to a stop, kinetic energy is transformed to select one:?

photoshop1234 [79]

Potential energy, I’m pretty sure I don’t know but we was learning this in science and this is all I remember that potential energy is the moment energy reaches to a stop...

7 0

3 years ago

Read 2 more answers

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0

lora16 [44]

Answer:

$d=0.165m$

Explanation:

Given

$m=0.25kg$ , $x_{1}=5cm*\frac{1m}{100cm}=0.05m$ , $x_{2}=4cm*\frac{1m}{100cm}=0.04m$ , $v=2\frac{rev}{s}$

The tension of the spring is

$F_{k}=K*x_{1}=m*g$

$K=\frac{m*g}{x_{1}}$

$K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m$

The force in the spring is equal to centripetal force so

$F_{c}=\frac{m*v^2}{r}$

$v=w*r=2\pi*r$

But Fc is also

Fc=KxΔr

$F_{c}=K*(r-x_{2})$

Replacing

$m*4\pi^2*r=K*(r-x_{2})$

$0.25kg*4\pi^2*r=49*(r-0.04m)$

$r=0.205m$

total distance is

$d=0.205-0.04=0.165m$

3 0

3 years ago

Calculate the force of a particle with a net charge of 170 coulombs traveling at a speed of 135 meters/second perpendicular to t

amm1812

Answer:

F=1.14N j

Explanation:

The magnitude of the magnetic force over a charge in a constant magnetic field is given by the formula:

$|\vec{F}|=|q\vec{v} \ X\ \vec{B}|=qvsin\theta$ (|)

In this case v and B vectors are perpendicular between them. Furthermore the direction of the magnetic force is:

-i X k = +j

Finally, by replacing in (1) we obtain:

$\vec{F}=(170C)(135\frac{m}{s})(5.0*10^{-5}T)=1.14N\ \hat{j}$

hope this helps!

6 0

3 years ago

Read 2 more answers