All categories

3 years ago

Practice

Physics

1 answer:

stiv31 [10]3 years ago

5 0

Answer:

The answer is Letter B The car travel at a constant veloc

You might be interested in

Help solve these two problems im having trouble trying to start these problems?

belka [17]

Answer:

25. Approximately 8.1 meters

26. North 1.31 km, and East 2.81 km

Explanation:

25.

Notice that the displacements: 6 meters east and 5.4 south create the legs of a right angle triangle. The hypotenuse of that triangle will be the distance (d) needed to cover in order to get the ball in the hole in one putt. That is:

$d=\sqrt{6^2+5.4^2} =\sqrt{65.16} \approx 8.072\,\,\,m$

which can be rounded to 8.1 m.

26.

Notice that the 3.1 km at an angle of 25 degrees north of east, is the hypotenuse of a right angle triangle that has for legs the east and north components of that distance.

We can find the leg corresponding to the east displacement using the cosine function (that relates adjacent side with hypotenuse):

$east\,\, comp=3.1 * cos(25^o)\approx 2.809\,\,km$

and we can calculate the north component using the sine function that relates the opposite side to the angle with the hypotenuse.

$north\,\,component = 3.1 * sin(25^o) \approx 1.31 \,\,km$

6 0

3 years ago

Which type of wire would have the least resistance?

telo118 [61]

The answer is E. Short and thick while cold.

4 0

4 years ago

Question 8 of 10 When is there the least amount of heat transfer within a liquid? O A. When the substance changes into a gas OB.

AlexFokin [52]

Answer:

C. When the temperature of the liquid is the same throughout

Explanation:

4 0

3 years ago

The acceleration due to gravity vector is always in the ____ direction.

finlep [7]

It goes in the downward direction

4 0

3 years ago

A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit

Ratling [72]

Answer:

The magnitude of angular acceleration is $232.38\ rad/s^2$ .

Explanation:

Given that,

Initial angular velocity, $\omega_i=3500\ rev/min=366.5\ rad/s$

When it switched off, it comes o rest, $\omega_f=0$

Number of revolution, $\theta=46=289.02\ rad$

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2$

So, the magnitude of angular acceleration is $232.38\ rad/s^2$ . Hence, this is the required solution.

6 0

3 years ago