Recording the location of a star requires a measurement of:

what is the magnitude of the vector described below? 13 m/s to the east a. east b. meters per second c. 13 m/s d. meters

Katen [24]

Answer:

C

Explanation:

Magnitude of any quantity is the measurable value of the quantity. While the direction of the given quantity is the specific pointing direction of position or the angle at which it move.

The magnitude of the vector described below? 13 m/s to the east will be 13 m/s

While the direction will be eastward.

Therefore, the magnitude is 13 m/s

The correct answer is option C

3 0

3 years ago

7) T F If two forces of equal magnitude act on an object that is hinged at a pivot, the force acting farther from the pivot must

saul85 [17]

Answer:

False

Explanation:

The torque exerted by a force is given by:

$\tau=Fd sin \theta$

where

F is the magnitude of the force

d is the distance between the point of application of the force and the pivot

$\theta$ is the angle between the directions of F and d

We see that the magnitude of the torque depends on 3 factors. In this problem, we have 2 forces of equal magnitude (so, equal F). Moreover, one of the forces (let's call it force 1) acts farther from the pivot than force 2, so we have

$d_1 > d_2$

However, this does not mean that force 1 produces a greater torque. In fact, it also depends on the angle at which the force is applied. For instance, if the first force is applied parallel to d, then we have

$\theta_1 =0\\sin \theta=0$

and the torque produced by this force would be zero.

So, the statement is false.

4 0

3 years ago

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T on Earth. Calculate the

melamori03 [73]

Answer:

T = T

Explanation:

Time period of a simple pendulum is not affected by the mass of the bob. As we know, $T = 2\pi \sqrt{L/g}$ . There is no factor of mass affecting when we derived the equation. The basic reason behind the time period is not affected is because of mass dependence on angular acceleration. As the mass increases the acceleration increase and the Time Period remains constant.

8 0

4 years ago

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

4 years ago

Two blocks, with masses M2>M1, are connected by ropes. You pull to the right on a second rope, with external force "T1".The b

Gre4nikov [31]

Answer:

$(M_1 + M_2) a > M_2 a$

Becuase $M_1 +M_2> M_2$

So then we can conclude that:

$T_1 > T_2$

And that makes sense since the force $T_1$ needs to accelerate the two masses and $T_2$ just need to accelerate $M_2$ .

So the best option for this case would be:

a. T1 > T2

See explanation below.

Explanation:

For this case we consider the system as shown on the figure attached.

Since the system is connected the acceleration for both masses are equal, that is $a_{M_1}= a_{M_2} = a$

From the second Law of Newthon we have that the force applied for the mass $M_2$ is $F_{M_2}= M_2 a$ and we know that the force acting on the x axis for the mass 2 is $F_{M_2}= T_2$ so then we have that $T_2= M_2 a$

Now when we consider the system of $M_1 +M_2$ as a whole mass, this system have the same acceleration $a$ and on this case we will see that the only force acting on the entire system would be $T_1$ and then by the second law of Newton we have that: