Ask question

Ask question

All categories

3 years ago

5

A truck is moving around a circular curve at a uniform velocity of 13 m/s. If the centripetal force on the truck is 3,300 N and

the mass of the truck is 1,600 kg, what's the radius of the curve?
A. 6.30 m
B. 81.94 m
C. 109.09 m
D. 348.56 m

2 answers:

Paladinen [302]3 years ago

4 0

Answer: Option B.

Since here the truck is moving on a circular track, it will experience centripetal force.

F(centripetal) = m × acc
or
$r = \frac{m v^{2}}{F}$

where r is the radius of the track.
m is the mass of truck
v is the speed of the truck.
Given: v = <span>13 m/s
m = </span><span>1,600 kg
</span>F = 3300 Newton

To find = radius of track=?
$r = \frac{m v^{2} }{F}$
$r = \frac{1600*13*13}{3300}$
r = 81.94 m
Therefore, radius of track is 81.94 m

I am Lyosha [343]3 years ago

3 0

The Correct answer to this question for Penn Foster Students is: 81.94m

You might be interested in

A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

3 years ago

Limestone formations in caves are considered what kind of weathering?

Zanzabum

Answer:

A.

Explanation:

because the carbonic acid reacts to the limestone.

8 0

2 years ago

An object accelerates 13.0 m/s2 when a force of 4.4 newtons is applied to it. What is the

velikii [3]

Answer:

0.3384 N

Explanation:

Acceleration = 13 m/s^2

Force = 4.4 N

Force = mass * acceleration

mass = force / acceleration

mass = 4.4 / 13

mass = 0.3384 N

6 0

2 years ago

A small bolt with a mass of 41.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire

EleoNora [17]

Answer:

Amplitude will be equal to 0.091 m

Explanation:

Given mass of the slits = 41 gram = 0.041 kg

Frequency f = 1.65 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 1.65=10.362rad/sec$

Angular frequency is equal to $\omega =\sqrt{\frac{k}{m}}$

$10.362 =\sqrt{\frac{k}{0.041}}$

Squaring both side

$107.371 ={\frac{k}{0.041}}$

k = 4.40 N/m

For vertical osculation

$mg=kA$

$0.041\times 9.8=4.40\times A$

A = 0.091 m

So amplitude will be equal to 0.0391 m

8 0

3 years ago

Physics Circuit, need a walkthrough

QveST [7]

I don't see a picture

7 0

2 years ago