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Sergio [31]
3 years ago
9

The force of gravity on Mercury is less than the Earth because:

Physics
2 answers:
drek231 [11]3 years ago
8 0
The answer is B.

Earth's mass is <span>5.972 × 10^24 kg
Mercury's mass is </span><span>3.285 × 10^23 kg

Hope i helped</span>
Vinvika [58]3 years ago
7 0
Mercury has less mass than earth. So the answer is B
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A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is
Umnica [9.8K]

Answer:

<h2>Magnitude of the second charge is -4.17*10^{-7}C</h2>

Explanation:

According to columbs law;

F = kq1q2/r^{2}

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

0.03=9*10^{9}*  -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C

6 0
4 years ago
Read 2 more answers
12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor = how
UkoKoshka [18]

Answer:

12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.

Explanation:

In the United States, a standard "drink" of beer has 12 ounces, a standard "drink" of wine has 5 ounces and standard drink of liquor has 1.5 ounces. Then, we obtain the quantity of drinks by dividing the total volume of each drink by its respective unit volume and summing each term. That is:

N = \frac{12\,oz}{12\,\frac{oz}{dr} }+\frac{12\,oz}{5\,\frac{oz}{dr} }+\frac{3\,oz}{1.5\,\frac{oz}{dr} }

N = 1\,dr+2.4\,dr+2\,dr

N = 5.4\,dr

N = 6\,dr

12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.

8 0
3 years ago
A 26-kg sled is on a snow-covered slope. The coefficients of friction between the sled’s runners and the snow are µs = 0.096 and µ
sweet-ann [11.9K]

Answer:

Explanation:

Given

mass of sled =26 kg

coefficient of static friction \mu _s=0.096

coefficient of kinetic friction \mu _k=0.072

In order to move sled from rest we need to provide a force greater than static friction which is given by

f_s=\mu mg=0.096\times 26\times 9.8=24.46 N

After Moving Sled kinetic friction comes in to play which is less than static friction

f_k=\mu _kmg=0.072\times 26\times 9.8=18.34 N

therefore minimum force to keep moving sledge at constant velocity is 18.34 N

3 0
3 years ago
The y component of the electric field of an electromagnetic wave traveling in the +x direction through vacuum obeys the equation
Alexxx [7]

Answer: 8.6 µm

Explanation:

At a long distance from the source, the components (the electric and magnetic fields) of the electromagnetic waves, behave like plane waves, so the equation for the y component of the electric field obeys an equation like this one:

Ey =Emax cos (kx-ωt)

So, we can write the following equality:

ω= 2.2 1014 rad/sec

The angular frequency and the linear frequency are related as follows:

f = ω/ 2π= 2.2 1014 / 2π (rad/sec) / rad = 0.35 1014 1/sec

In an electromagnetic wave propagating through vacuum, the speed of the wave is just the speed of light, c.

The wavelength, speed and frequency, are related by this equation:

λ = c/f

λ = 3.108 m/s / 0.35. 1014 1/s = 8.6 µm.

7 0
3 years ago
4. Think back to Coulomb's Law. Two coins with identical charges are placed on a lab table 1.35 m apart.
FinnZ [79.3K]
A) To calculate the charge of each coin, we must apply the expression of the Coulomb's Law:
 
 F=K(q1xq2)/r²
 
 F: The magnitud of the force between the charges. (F=2.0 N).
 K: Constant of proporcionality of the Coulomb's Law (K=9x10^9 Nxm²/C²).
 q1 and q2: Electrical charges.
 r: The distance between the charges (r=1.35 m).
 
 We have the values of F, K and r, so we can calculate q1xq2, because both<span> coins have  identical charges:
</span> 
 q1xq2=(r²xF)/K
 q1xq2=(1.35 m)²(2.0 N)/9x10^9 Nxm²/C²
 q1xq2=3x10^-10 C
 q1=q2=(<span>3x10^-10 C)/2
 
 </span>Then, the charge of each coin, is:
<span> 
 q1=1.5x</span><span>10^-10 C 
 
 </span>q2=1.5x10^-10 C

B) <span>Would the force be classified as a force of attraction or repulsion?
</span> 
 It is a force of repulsion, because both coins have identical charges and both are postive. In others words, when two bodies have identical charges (positive charges or negative charges), the force is of repulsion.
5 0
3 years ago
Read 2 more answers
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