3 years ago

One of the most effective ways to evaluate data is to try to replicate it.

Physics

1 answer:

zubka84 [21]3 years ago

6 0

I believe the answer is False

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A 5.5 kg box is pushed across the lunch table. The net force applied to the box is 9.7 N. What is the acceleration of the box?

Marysya12 [62]

Answer:

F=ma

a=F/m=9.7N/5.5Kg=1.8 m/s^2

3 0

3 years ago

What is the size of the force acting on a copper wire with a magnetic flux density of 3.6 x 10-2 T acting at

anyanavicka [17]

Answer:

F = 0.414 N

Explanation:

Given that,

Magnetic flux density, $B=3.6\times 10^{-2}\ T$

The length of the wire, l = 24 m

Current, I = 0.48 A

We need to find the force acting on the wire. The formula for the force is given by:

$F=ILB$

Put all the values,

$F=0.48\times 24\times 3.6\times 10^{-2}\\\\F=0.414\ N$

So, the force acting on the copper wire is equal to 0.414 N.

7 0

3 years ago

Determine the ratio of the electrostatic force to the gravitational force between a proton and an electron, FE/FG. Note: k = 8.9

Paraphin [41]

Answer:

$\frac{F_e}{F_g} = 2.3 \times 10^{18}$

Explanation:

The gravitational force is given by Newton's Law of Gravity:

$F_g = \frac{Gm_1m_2}{r^2}$

The electrostatic force is given by Coulomb's Law:

$F_e = \frac{kq_1q_2}{r^2}$

The ratio between these two forces is

$\frac{F_e}{F_g} = \frac{\frac{kq_1q_2}{r^2}}{\frac{Gm_1m_2}{r^2}} = \frac{kq_1q_2}{Gm_1m_2} = \frac{(8.99\times 10^{-12})(1.6\times 10^{-19})(1.6 \times 10^{-19})}{(6.67\times 10^{-11})(9.1 \times 10^{-31})(1.6 \times 10^{-27})} = 2.3 \times 10^{18}$