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3 years ago

What is the motion of the object?

Physics

1 answer:

aleksley [76]3 years ago

4 0

Answer:

<em>Thus, the object is accelerating to the left</em>

Explanation:

<u>The Net Force</u>

The net force is the result of adding all the forces as vectors acting on a body.

$\vec F=\vec F_1+\vec F_2+...+\vec F_n$

Each vector can be expressed in its rectangular components Fx and Fy, and the sum is the sum of the rectangular components separately.

Second Newton's law gives the relation between the net force and the acceleration of the body:

$\vec F = m.\vec a$

We can see the acceleration is a vector with the same direction as the net force.

The diagram shows two vertical forces and two horizontal forces.

The vertical forces are acting in opposite directions and with the same magnitude, thus they cancel out, leaving zero net force in the y-axis.

The horizontal forces are opposite and with different magnitudes. Since the force acting to the left (F3) has a greater magnitude than the force acting to the right (F4), there is a net force directed to the left with a magnitude of 60 N - 20 N = 40 N

Thus, the object is accelerating to the left

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A woman throws a javelin 35 mph at an angle 30 degrees from the ground. Neglecting wind resistance or the height the javelin thr

Anna71 [15]

Answer:

35 mph

Explanation:

The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.

When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.

When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.

So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.

5 0

2 years ago

A car with a total mass of 1800 kg (including passengers) is driving down a washboard road with bumps spaced 4.9 m apart. The ri

Drupady [299]

Answer:

k = 9.6 x 10^5 N/m or 9.6 kN/m

Explanation:

First, we need to use the expression to calculate the spring constant which is:

w² = k/m

Solving for k:

k = w²*m

To get the angular velocity:

w = 2πf

The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:

f = V/x

f = 5.7 / 4.9 = 1.16 Hz

Now the angular velocity:

w = 2π*1.16

w = 7.29 rad/s

Finally, solving for k:

k = (7.29)² * 1800

k = 95,659.38 N/m

In two significant figures it'll ve 9.6 kN/m

5 0

2 years ago

Read 2 more answers

The period (T) of an oscillating wave is 1/5s. What happens to the frequency (f) of the wave if T increases to 1/2s

Anastasy [175]

Frequency = 1/T
as the 5 is reduced, frequency is increase.
as 1 whole wave travels through a point in a lesser time now

6 0

2 years ago

What is the direction of the force on a positive charge when passing through a magnetic field as indicated in this diagram? Expl

maria [59]

Answer:

The direction of the magnetic force on a moving charge is perpendicular to the plane formed by v and B and follows right hand rule–1 (RHR-1)

Explanation:

hope this helps

3 0

3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

3 years ago