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2 years ago

Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.

Physics

1 answer:

Rufina [12.5K]2 years ago

4 0

Answer: 1. decreasing the mass of both objects

2. decreasing the mass of one of the objects

3. increasing the distance between the objects

Explanation: Hope that helped! (:

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What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 1014 s−1 ejects a photon with 7.74 × 10−20 J

FrozenT [24]

Answer:

5.5 × 10^14 Hz or s^-1

no orange light has less frequency so no photoelectric effect

Explanation:

hf = hf0 + K.E

HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s

f is frequency of incident photon and f0 is threshold frequency

hf0 = hf- k.E

6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20

6.63 × 10 ^-34 × f0 = 3.64158×10^-19

f0 = 3.64158×10^-19/ 6.63 × 10 ^-34

f0 = 5.4925 × 10^14

f0 =5.5 × 10^14 Hz or s^-1

frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light

8 0

2 years ago

A na+ ion moves from inside a cell, where the electric potential is -72 mv, to outside the cell, where the potential is 0 v. wha

Vlada [557]

The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:
$\Delta U = q \Delta V = q (V_f - V_i)$
where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge: $q=1.6 \cdot 10^{-19} C$ , and Vf and Vi are the final and initial voltages.

Substituting the numbers, we find:
$\Delta U = (1.6 \cdot 10^{-19}C)(0 V-(-0.072 V))=1.15 \cdot 10^{-20} J$

7 0

2 years ago

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An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's

ivolga24 [154]

1) Focal length

We can find the focal length of the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$ (1)
where
f is the focal length
$d_o$ is the distance of the object from the mirror
$d_i$ is the distance of the image from the mirror

In this case, $d_o = 18 cm$ , while $d_i=-4 cm$ (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
$\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}$
from which we find
$f=- \frac{36}{7} cm=-5.1 cm$

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
$r=2f=2 \cdot 5.1 cm=10.2 cm$

3 0

2 years ago

Cassy shoots a large marble (Marble A, mass: 0.06 kg) at a smaller marble (Marble B, mass: 0.03 kg) that is sitting still. Marbl

Lostsunrise [7]

Conservation of linear momentum:

m*v inital = m*v final

0.06*0.7 + 0.03*0 = 0.06*(-0.2) + 0.03*v

(my algebra, or use ur calculator: 0.06*.07=0.042, etc ... or ur teacher may think you got some help)

0.06*(0.7+0.2)=0.03*v, v = 0.06*0.9/0.03=1.8 m/s

Answer 1.8 m/s (positive, to the right).

4 0

2 years ago

Joe learned that decaying plants produce an acidic environment. Joe wants to know if pill bugs prefer acidic environments instea

Anna11 [10]

Yessgshhshe the answer

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2 years ago

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