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zavuch27 [327]
3 years ago
9

Consider the gravity of the four spheres. If we wanted sphere C to attract the other three spheres, what is one feature of spher

e C we must change?
A) Increase its mass.
B) Make it less dense.
C) Increase its diameter.
D) Change its composition.
Physics
2 answers:
Diano4ka-milaya [45]3 years ago
6 0

Answer:

B

Explanation:

otez555 [7]3 years ago
4 0

Answer:

The correct answer is A.

Explanation:

A greater mass means a greater gravitational pull. To give Sphere C more gravitational pull you mus increase its mass

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A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t
andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = \frac{r}{2}

f = \frac{0.40}{2} = 0.20 m

Now,

m = - \frac{v}{u}

- 2 = -\frac{v}{u}

\frac{v}{u} = 2                  (2)

Now, by lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

v = \frac{uf}{u - f}            (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = \frac{uf}{u - f}

2 = \frac{f}{u - f}

2(u - 0.20) = 0.20

u = 0.30 m

6 0
3 years ago
The uncertainty in the position of an electron along an x axis is given as 5 x 10-12 m. What is the least uncertainty in any sim
Vsevolod [243]

Answer:

The least uncertainty in the momentum component px is 1 × 10⁻²³ kg.m.s⁻¹.

Explanation:

According to Heisenberg's uncertainty principle, the uncertainty in the position of an electron (σx) and the uncertainty in its linear momentum (σpx) are complementary variables and are related through the following expression.

σx . σpx ≥ h/4π

where,

h is the Planck´s constant

If σx = 5 × 10⁻¹²m,

5 × 10⁻¹²m . σpx ≥ 6.63 × 10⁻³⁴ kg.m².s⁻¹/4π

σpx ≥ 1 × 10⁻²³ kg.m.s⁻¹

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3 years ago
A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torqu
ira [324]

Answer:

-20.0 kg m^2/s

Explanation:

The angular momentum of an object in rotation is given by

L=I \omega

where

I is the moment of inertia

\omega is the angular speed

In this problem, initially we have

I=2 kg m^2 is the moment of inertia of the wheel

\omega_i = 6.0 rad/s is the initial angular speed

So the initial angular momentum is

L_i = I\omega_i = (2)(6.0)=12 kg m^2/s

Later, a counterclockwise torque of

\tau=-5.0 Nm is applied

So the angular acceleration of the wheel is:

\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2 in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

\omega_f = \omega_i + \alpha t

where

t = 4.0 is the time interval

Solving,

\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s

So, the change in angular momentum is:

\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2

7 0
3 years ago
Why would a flare be observed in visible light, when they are so much brighter in X-ray and ultraviolet light?
raketka [301]

Answer:

Because it radiates a spectre of frequencies, this means that it radiates in a continuous range of frequencies, ones with more intensity (like x-ray and ultraviolet) and others with less intensity (like the visible light). While most of the radiation is not in the visible range, there still is a part of the radiation in the visible spectre. And while this part is not the most intense part, the radiation is so large that we can see a very bright visible light.

3 0
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Damm [24]

Answer:

Stayed the same

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Ep = m*g*h

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