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Margaret [11]
2 years ago
8

1.

Physics
1 answer:
zubka84 [21]2 years ago
7 0

Answer:

D

Explanation:

Momentum= mass x speed (P=MV)

Momentum= 5x6=30

Momentum= 30kg m/s

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An 800 N man climbs 5 m up a ladder. How much gravitational potential energy does he gain?
Artemon [7]

Answer:

4000J

Explanation:

Given parameters:

Weight of the man  = 800N

Height of ladder  = 5m

Unknown:

Gravitational potential energy gained  = ?

Solution:

The gravitational potential energy is due to the position of a body.

 Gravitational potential energy = weight x height

Now insert the parameters;

 Gravitational potential energy  = 800 x 5  = 4000J

5 0
3 years ago
Can an ordinary object, like a motorcycle, be mass-less? Yes or No
Drupady [299]

Answer:

no.

Explanation:

because the mass of an object never changes.

4 0
3 years ago
IF THERE ARE ONLY 118 ELEMENTS, HOW DO YOU ACCOUNT FOR THE MANY MILLIONS OF THINGS THAT WE HAVE IN OUR UNIVERSE?
bogdanovich [222]

Answer:

theres only 118 elements that are discovered. now that they're the only ones out there

Explanation:

3 0
3 years ago
PLEASE HELP Due today!
BigorU [14]
So i believe is exercise:)
7 0
3 years ago
Two long parallel wires carry currents of 20 a and 5.0 a in opposite directions. the wires are separated by 0.20 m. what is the
Alex777 [14]
The two wires carry current in opposite directions: this means that if we see them from above, the magnetic field generated by one wire is clock-wise, while the magnetic field generated by the other wire is anti-clockwise. Therefore, if we take a point midway between the two wires, the resultant magnetic field at this point is just the sum of the two magnetic fields, since they act in the same direction.

Therefore, we should calculate the magnetic field generated by each wire and then calculate their sum. We are located at a distance r=0.10 m from each wire. 

The magnetic field generated by wire 1 is:
B_1= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(20 A)}{2 \pi (0.10 m)}=  4 \cdot 10^{-5}T

The magnetic field generated by wire 2 is:
B_2= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(5.0 A)}{2 \pi (0.10 m)}= 1 \cdot 10^{-5}T

And so, the resultant magnetic field at the point midway between the two wires is
B=B_1 + B_2 = 4 \cdot 10^{-5} T + 1 \cdot 10^{-5}T=5 \cdot 10^{-5} T
8 0
4 years ago
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