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Sav [38]
3 years ago
11

When stopping at a railroad crossing, you should be no closer than?

Physics
1 answer:
emmainna [20.7K]3 years ago
7 0
You should stop<span> before </span>crossing railroad<span> tracks: Whenever a </span>crossing<span> is not ... Follow </span>no closer than<span> 10 feet behind the large truck. </span>
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The maximum force that a grocery bag can withstand without ripping is 250 n. Suppose that the bag is filled with 20 kg of grocer
zhannawk [14.2K]

Answer:

Groceries stay in the bag.

Explanation:

Given:

Maximum force = 250 N

Bag filled with = 20 kg

Lifted acceleration = 5.0\ m/s^2

Solution:

We need to calculate the exerted force on the grocery bag by using Newton's second law.

F = ma

Where:

F = Exerted force on the object.

m = Mass of the object in kg

a = Acceleration of the object in m/s^2

Now, we substitute m = 20 kg and a = 5.0\ m/s^2 in Newton's second law,

F = 20\times 5.0

F = 100\ m/s^2

Since, the exerted force on the bag is less than 250 N, the groceries will stay in the bag.

3 0
3 years ago
Test questions!!!!!!!!
Pani-rosa [81]

Answer:

B. surface wave

8 0
2 years ago
A container is filled to a depth of 19.0 cm with water. On top of the water floats a 31.0-cm-thick layer of oil with specific gr
IceJOKER [234]

Answer:

Absolute pressure of the oil will be 102822.8 Pa  

Explanation:

We have given height h = 31 cm = 0.31 m

Acceleration due to gravity g=9.8m/sec^2

Specific gravity of oil = 0.600

So density of oil \rho =0.6\times 1000=600kg/m^3

We know that absolute pressure is given by P=P_0+\rho gh, here P_0=1.01\times 10^5Pa

So absolute pressure will be equal to P=1.01\times 10^5+600\times 9.8\times 0.31=102822.8Pa

So absolute pressure of the oil will be 102822.8 Pa

6 0
3 years ago
An attractive force of 7.2 N occurs between two point charges that are 0.10 m apart. If one charge is -4.0 µC, what is the other
Umnica [9.8K]
The answer is c. +2.0 µC

To calculate this, we will use Coulomb's Law:

F = k*Q1*Q2/r²

where F is force, k is constant, Q is a charge, r is a distance between charges.

k = 9.0 × 10⁹  N*m/C²


It is given:

F = 7.2 N

d = 0.1 m =  10⁻¹ m

Q1 = -4.0 µC = 4 * 1.0 × 10⁻⁶ = 4.0 × 10⁻⁶

Q2 = ?


Thus, let's replace this in the formula for the force:

7.2 = 9.0 × 10⁹ * 4.0 × 10⁻⁶ * Q2/(10⁻¹)²

7.2 = 9 * 4 * 10⁹⁻⁶ * Q2/10⁻¹°²

7.2 = 36 × 10³ * Q2 / 10⁻²

Multiply both sides of the equation by 10⁻²:

7.2 × 10⁻² = 36 × 10³ * Q2

⇒ Q2 = 7.2 × 10⁻² / 36 × 10³ = 7.2/36 × 10⁻²⁻³ = 0.2 × 10⁻⁵ = 2 × 10⁻⁶ 


Since µC = 1.0 × 10^-6:

Q2 = 2 * 1.0 × 10^-6 = 2 µC

5 0
3 years ago
Q1. After three half-lives of an isotope, 1 billion of the original isotope's atoms still remain in a certain amount of this ele
77julia77 [94]
 <span>If 1 eighth equals 1 billion 7 eighth equals 7 billion. 

The asker of the second question needs a tutorial in radiometric dating. There is little likelihood that the daughter isotope has the same atomic weight as the parent isotope. To measure the mass isotopes doesn't tell us how many atoms of each exist. To get around that let's pretend — which will likely serve the purpose ineptly intended — that the values give an the particle ratio, 125:875. 

The original parent isotope count was 125 + 875 = 1000. The remaining parent isotope is 125/1000 or 1/8. 1/8 = (1/2)^h, where h is the number of half-lives. 

h = log (1/8) ÷ log(1/2) = 3 

And 3 half-lives • 150,000 years/half-life = 450,000 years.</span>
5 0
3 years ago
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