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3 years ago

The forces exerted on an object are shown. (3 points)

Physics

1 answer:

Alex_Xolod [135]3 years ago

7 0

Answer:

<em>F equals 3 N and the object remains stationary</em>. (second option in the list)

Explanation:

For sure to cancel acting forces, F must be 3N pointing up. But with regards to the object stationary or not, the question is tricky. We could have a ZERO net force applied, and the object moving at constant speed, which could still verify Newton's Laws. But considering the first answer option that refers to vertical motion upward where the object could be gaining potential energy, the most accurate response is that the force F has to be 3 N pointing up to make the object in equilibrium, and no motion in the vertical axis.

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How long does it take an automobile traveling 66.7 km/h to become even with a car that is traveling in another lane at 52.7 km/h

tresset_1 [31]

Answer:

The time taken is $t = 32.5 \ s$

Explanation:

From the question we are told that

The speed of first car is $v_1 = 66.7 \ km/h = 18.3 \ m/s$

The speed of second car is $v_2 = 52.7 \ km/h = 14.64 \ m/s$

The initial distance of separation is $d = 119 \ m$

The distance covered by first car is mathematically represented as

$d_t = d_i + d_f$

Here $d_i$ is the initial distance which is 0 m/s

and $d_f$ is the final distance covered which is evaluated as $d_f = v_1 * t$

$d_t = 0 \ m/s + (v_1 * t )$

$d_t = 0 \ m/s + (18.3 * t )$

The distance covered by second car is mathematically represented as

$d_t = d_i + d_f$

Here $d_i$ is the initial distance which is 119 m

and $d_f$ is the final distance covered which is evaluated as $d_f = v_2* t$

$d_t = 119 + 14.64 * t$

Given that the two car are now in the same position we have that

$119 + 14.64 * t = 0 + (18.3 * t )$

$t = 32.5 \ s$

6 0

3 years ago

The Andromeda galaxy is the closest major galaxy to our own. Andromeda shows a distinct blue-shift of light when we analyze it.

saul85 [17]

At the present time, the only way we know of that light can get shifted
toward the blue end of the spectrum is the Doppler effect ... wavelengths
appear shorter than they should be when the source is moving toward us.

IF that's true in the case of the Andromeda galaxy, it means the galaxy is
moving toward us.

We use the same reasoning to conclude that all the galaxies whose light is red-shifted are moving away from us. That includes the vast majority of all galaxies that we can see, and it strongly supports the theory of the big bang
and the expanding universe.

If somebody ever comes along and discovers a DIFFERENT way that light
can get shifted to new, longer or shorter wavelengths, then pretty much all
of modern Cosmology will be out the window. There's a lot riding on the
Doppler effect !

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3 years ago

The measurement of an exoplanet's radius is measured in units compared to ________.

sesenic [268]

The eaths radius is the correct answer, if you need proof look at nasa's website

4 0

3 years ago

Read 2 more answers

An ambulance driver traveling at 31.0 m/s (69.3 mph) honks his horn as he sees a motorist ahead on the highway traveling in the

IrinaVladis [17]

Answer:

fo = 378.52Hz

Explanation:

Using Doppler effect formula:

$f'=\frac{C-Vb}{C-Va}*fo$

where

f' = 392 Hz

C = 340m/s

Vb = 20m/s

Va = 31m/s

Replacing these values and solving for fo:

fo = 378.52Hz

4 0

3 years ago

Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s

anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of $-5.2^{\circ}$

D) The impulse exerted on C is 4.29 kg m/s at a direction of $-5.2^{\circ}$

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

$p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B$

$m_A = 0.600 kg$ is the mass of sphere A

$u_A = 4.00 m/s$ is the initial velocity of the sphere A (taking the right as positive direction)

$v_A$ is the final velocity of sphere A

$m_B = 1.80 kg$ is the mass of sphere B

$u_B = 2.00 m/s$ is the initial velocity of the sphere B

$v_B = 3.00 m/s$ is the final velocity of the sphere B

Solving for vA:

$v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s$

The sign is positive, so the direction is to the right.

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

$K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J$

After the collision:

$K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J$

The total kinetic energy is conserved: therefore, the collision is elastic.

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

$p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s$

While the total momentum along the y-axis is zero:

$p_y = 0$

We can now write the equations of conservation of momentum along the two directions as follows:

$p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy}$ (1)

We also know the components of the momentum of B after the collision:

$p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s$

So substituting into (1), we find the components of the momentum of C after the collision:

$p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s$

So the magnitude of the momentum of C is

$p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s$

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

$v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s$

And the direction is

$\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}$

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

$p_C = 0$

While the final momentum is:

$p'_C = 4.29 kg m/s$

So the magnitude of the impulse exerted on C is

$I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s$

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore $-5.2^{\circ}$ .

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

$K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J$

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

$K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J$

Since the total kinetic energy is not conserved, the collision is inelastic.

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

$F=Ma$

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

brainly.com/question/6439920

brainly.com/question/2990238

brainly.com/question/7973509

brainly.com/question/6573742

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3 years ago