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3 years ago

After their loss to U of A, the ASU sumdevils traveled back to Tempe which is 187

Physics

1 answer:

Dmitry_Shevchenko [17]3 years ago

5 0

The average speed is 20.8 m/s

Explanation:

The average speed for the trip is given by:

$v=\frac{d}{t}$

where

d is the distance covered

t is the time elapsed

For the trip in this problem, we have:

d = 187 km = 187,000 m is the distance travelled

The initial time is 10:00 pm while the arriving time is 12:30 am: this means that the time elapsed is 2.5 hours. Converting into seconds,

$t = 2.5 h \cdot (60)(60)=9000 s$

Therefore, the average speed for the trip is

$v=\frac{187,000}{9000}=20.8 m/s$

Learn more about average speed:

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An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas

Viefleur [7K]

1) Available force of friction: 6174 N

2) No

Explanation:

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

$F_f=\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

$\mu=0.500$ (coefficient of friction)

m = 1260 kg (mass of the car)

$g=9.8 m/s^2$

Therefore, the force of friction is

$F_f=(0.500)(1260)(9.8)=6174 N$

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

$F=m\frac{v^2}{r}$

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

$v=54.1 km/h =15.0 m/s$ is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

$F=(1260)\frac{15.0^2}{41.6}=6814 N$

Therefore, the force of friction is not enough to keep the car in the curve, since $F_f$

4 0

3 years ago

________ occurs when an electric current is made by a changing magnetic field

Butoxors [25]

Induced electromotive force

4 0

3 years ago

A truck travels to and from a stone quarry that is located 2.5km to the east.

Harrizon [31]

Answer:

Distance is 5.0 km and displacement is 0 km.

Explanation:

The diagram is shown below.

Distance is the total length that the truck covers along the complete journey. The truck moves from a stone quarry covering a length of 2.5 km and then again come back to the same point and thus covering another 2.5 km. Therefore, the total length covered by the truck is the sum of forward and backward paths which is equal to 5.0 km.

Now, displacement depends only on the initial and final position. It is given as the difference of final position and initial position. So, the truck starts from stone quarry and returns back to the same position. So, the initial position and final position is same. Therefore, the difference is 0 and hence displacement is zero.

5 0

3 years ago

A grating that has 3900 slits per cm produces a third- order fringe at a 28.0° angle. What wavelength of light is being used? Ex

Zolol [24]

Explanation:

Given that,

Number of slits per cm, $N=3900\ lines/cm=390000\lines/m$

The third fringe is obtained at an angle of, $\theta=28$

We need to find the wavelength of light used. The grating equation is given by :

$d\ sin\theta=n\lambda$

$d=\dfrac{1}{N}=\dfrac{1}{390000}$

$d=2.56\times 10^{-6}\ m$

$\lambda=\dfrac{d\ sin\theta}{n}$ , n = 3

$\lambda=\dfrac{2.56\times 10^{-6}\times \ sin(28)}{3}$

$\lambda=4.006\times 10^{-7}\ m$

$\lambda=400\ nm$

So, the wavelength of the light is 400 nm. Hence, this is the required solution.

4 0

3 years ago

PIUDICITIS CONSECulvely and Circle your aliswers. Lilyo

mihalych1998 [28]

Answer:

The angular acceleration of CD player is $13.93\ rad/s^2$ .

Explanation:

Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.

It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :

$a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2$

So, the angular acceleration of CD player is $13.93\ rad/s^2$ .

7 0

2 years ago