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Leya [2.2K]
3 years ago
15

When a constant force acts on an object, what does the object's change in momentum depend upon?

Physics
1 answer:
Eva8 [605]3 years ago
8 0
You don’t need time (1) or the force (4) either so just 3,5
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An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th
kipiarov [429]

Answer:

2.24 \times 10^{-14} Newtons

Explanation:

Magnitude of charge on the electron = q = 1.6 \times 10^{-19} Coulombs

The negative sign in the question statement indicates that the charge is negative.

Magnitude of Electric Field experienced by the electron = E = 1.4 \times 10^{5} Newtons/Coulomb

Magnitude of Force on the electron = F = ?

The relation between the charge, electric field and the force on the charge because of electric field is given by:

E=\frac{F}{q}

From here we can write:

F = qE

Using the values, we get:

F = 1.6 \times 10^{-19} \times 1.4 \times 10^{5}\\\\ F = 2.24 \times 10^{-14}

Thus, the magnitude of the electric force experience by the electron would be 2.24 \times 10^{-14} Newtons

8 0
3 years ago
How many significant digits are in the following measurements?<br> a. 1300 m
Fofino [41]

Answer:

For example, 1300 with a bar placed over the first 0 would have three significant figures (with the bar indicating that the number is precise to the nearest ten).

Explanation:

hope it helps :)

5 0
2 years ago
I do not know where to start.
irakobra [83]
Underline the words then eliminate the ones that arent part of the problem!
4 0
3 years ago
What energy transfer will a stretched rubber band have when let go
GarryVolchara [31]

Answer:

when the rubber band is realeased the potential energy is quickly converted to kinetic energy this is equal to one mass of the the rubber band multiplied by its velocity( in meters per second)

3 0
3 years ago
A "gauge 8" jumper cable has a diameter d of 0.326 centimeters. The cable carries a current I of 30.0 amperes. The electric fiel
AveGali [126]

Answer:

0.0979 N/c

Explanation:

Electric field, E is given as a product of resistivity and current density

E=jP where P is resistivity and j is current density

But the current density is given as

j=\frac {I}{A} where I is current and A is area and A=\pi r^{2}

Substituting this into the first equation then E=P\times \frac {I}{\pi r^{2}}

Given diameter of 0.259 cm= 0.00259 m and the radius will be half of it which is 0.001295 m

E=1.72\times 10^{-8}\times \frac {30}{\pi \times 0.001295^{2}}=9.79\times 10^{-2} N/c=0.0979 N/c

4 0
3 years ago
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