two teams are playing tug of war team a pulls to the right with a force of 450N Team B pulls to the left with a force of 415 N

The electric current in a wire is 1.5A. How many electrons flow past a given point in a time of 2s?

kipiarov [429]

Answer:

The amount of electrons that flow in the given time is 3.0 C.

Explanation:

An electric current is defined as the ratio of the quantity of charge flowing through a conductor to the time taken.

i.e I = $\frac{Q}{t}$ ...................(1)

It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.

In the given question, I = 1.5A, t = 2s, find Q.

From equation 1,

Q = I × t

= 1.5 × 2

= 3.0 Coulombs

The amount of electrons that flow in the given time is 3.0 C.

5 0

3 years ago

Anakin Skywalker's pod racer has a mass of 450 kg. If the top speed of this racer is 947

andriy [413]

Answer:

15,569,653.3 Joules(J)

Explanation:

The equation used to find Kinetic Energy (KE) is

KE = $\frac{1}{2}$ m $v^{2}$

You have been given

m = 450kg

v = 947km/h

KE = ???

Firstly, we need to convert the km/h into m/s as this is the unit used in the KE equation

This can be done by dividing by 3.6

947km/h = 263.056m/s

Substitute you values into the equation

KE = $\frac{1}{2}$ m $v^{2}$

KE = $\frac{1}{2}$ * 450 * $263.056^{2}$

KE = 15,569,653.3 Joules(J)

Round your answer as appropriate

3 0

3 years ago

In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a

SOVA2 [1]

Answer: $0.084 m/s^2$

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance $d_1=8.13 km$

time taken=25 min =1500 s

initial speed $v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s$

after 8.13 km mark steve started to accelerate

speed after 60 s

$v_2=v_1+at$

$v_2=5.6+a\times 60$

distance traveled in 60 sec

$d_2=v_1\times 60+\frac{a60^2}{2}$

$d_2=336+1800 a$

time taken in last part of journey

$t_3=1663.6-1560=103.6 s$

distance traveled in this time

$d_3=v_2\times t_3$

$d_3=\left ( 5.6+a\times 60\right )103.6$

and total distance $=d_1+d_2+d_3$

$10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$a=0.084 m/s^2$

5 0

4 years ago

A. Test used to measure flexibility B. The elasticity of muscle groups C. Test approve by the president of the United States D.

olchik [2.2K]

1. C
2. G 
3. H 
4. J 
5. B 
6. I 
7. D 
8. E 
9. A 
10. F and For the best answers, search on this site https://shorturl.im/FbQuG

4 0

3 years ago

What aspect of motion can you conclude is common among freely falling objects ?

Liula [17]

Answer:

Free-fall is defined as the movement where the only force acting on an object is the gravitational force.

By the second Newton's law, we have that:

F = m*a

Where F = Force, m = mass, a = acceleration.

We can write this as:

a = F/m

And the gravitational force can be written as:

F = (G*M/r^2)*m

Where G is the gravitational constant, M is the mass of the Earth in this case, and r is the distance between both objects (the center of the Earth and the free-falling object)

As the radius of the Earth is really big, the term inside the parentheses is almost constant in the region of interest, then we can write:

G*M/r^2 ≈ g

And the gravitational force is:

F = g*m

And by the second Newton's law we had:

a = F/m = (g*m)/m = g

a = g

Then the acceleration does not depend on the mass of the object.

Then the thing that is common among the free-falling objects is the vertical acceleration.

6 0

3 years ago