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Jet001 [13]
3 years ago
12

How long will it take to go 150 km [E] traveling at 50 km/hr?

Physics
2 answers:
pantera1 [17]3 years ago
7 0
It will take 3 hours
Anit [1.1K]3 years ago
7 0

Answer:

SHTUP PErIODT POOH

Explanation:

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125 cm of gas are collected at 15 °C and
aleksklad [387]
The new volume will be ≈26mL
rounded to two significant figures.
Explanation:
This question involves the combined gas law. The equation to use is:
When working with gas laws, the temperature is always in Kelvins. To get Kelvins from a Celsius temperature, add 273.15 to the Celsius temperature.
6 0
3 years ago
Consider two conducting spheres with one having a larger radius than the other. Both spheres carry the same amount of excess cha
RSB [31]

Answer: Option (a) is the correct answer.

Explanation:

It is known that potential energy is the energy occupied by an object or substance due to its position is known as potential energy.

Therefore, more is the space occupied by an object more will be its position at a particular location. Hence, more will be its potential energy. On the other hand, smaller is the space occupied by an object, smaller will be the position holded by it.

Hence, smaller will be its potential energy.

Thus, we can conclude that for the given situation the statement, potential energy of the larger sphere is greater than that of the smaller sphere, is true.

6 0
3 years ago
Calculate the mass 9f the earth, assuring that uts is sphere with radius 6.67×10^6m.​
MA_775_DIABLO [31]

Answer:

6.86 × 10²⁴ kg

Explanation:

The mass of the earth m = density of earth, ρ × volume of earth, V

m = ρV

The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m

Since m = ρV

m = ρ4πr³/3

So, substituting the values of the variables into the equation for the mass of the earth, m, we have

m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3

m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3

m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3

m = 6546105.64378π × 10¹⁸ kg/3

m = 20565197.400122 × 10¹⁸ kg/3

m = 6855065.8 × 10¹⁸ kg

m = 6.8550658 × 10²⁴ kg

m ≅ 6.86 × 10²⁴ kg

8 0
3 years ago
Two point charges each have a value of 3.0 c and are separated by a distance of 4.0 m. what is the electric field at a point mid
swat32
 <span>Place a test charge in the middle. It is 2cm away from each charge. 
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point. 
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out. 
THIS IS A TRICK QUESTION. 
THe electric field exactly midway between them = 0/Q = 0. 
But if the point moves even slightly you need the following formula 
F= (1/4Piε)(Q1Q2/D^2) 
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
4 0
3 years ago
A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 ms from an initial spe
dezoksy [38]

Answer:

the magnitude of the average contact force exerted on the leg is  3466.98 N

Explanation:

Given the data in the question;

Initial velocity of hand v₀ = 5.25 m/s

final velocity of hand v = 0 m/s

time interval t = 2.65 ms = 0.00265 s

mass of hand m = 1.75 kg

We calculate force on the hand F_{hand

using equation for impulse in momentum

F_{hand × t = m( v - v₀ )

we substitute

F_{hand × 0.00265  = 1.75( 0 - 5.25 )

F_{hand × 0.00265  = 1.75( - 5.25 )

F_{hand × 0.00265  = -9.1875

F_{hand = -9.1875 / 0.00265  

F_{hand = -3466.98 N

Next we determine force on the leg F_{leg

Using Newton's third law of motion

for every action, there is an equal opposite reaction;

so, F_{leg = - F_{hand

we substitute

F_{leg = - ( -3466.98 N )

F_{leg = 3466.98 N

Therefore, the magnitude of the average contact force exerted on the leg is  3466.98 N

5 0
3 years ago
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