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Studentka2010 [4]
4 years ago
10

How much force would it take to make a 60 kg object accelerate at 5 m/s/s?

Physics
2 answers:
mote1985 [20]4 years ago
5 0

Answer:

300 N

Explanation:

force = mass × acceleration

= 60 kg × 5m/s^2

= 300 N

murzikaleks [220]4 years ago
4 0
The Answer Would Be 300 N
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When you go for a walk which of the following forces is paired with the force of friction on your shoe
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Answer:

static

Explanation:

static friction pushes in the direction you are walking.

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Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of
STatiana [176]

Answer:

2.27\cdot 10^{49}

Explanation:

The gravitational force between the proton and the electron is given by

F_G=G\frac{m_p m_e}{r^2}

where

G is the gravitational constant

m_p is the proton mass

m_e is the electron mass

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N

The electrical force between the proton and the electron is given by

F_E=k\frac{q_p q_e}{r^2}

where

k is the Coulomb constant

q_p = q_e = q is the elementary charge (charge of the proton and of the electron)

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N

So, the ratio of the electrical force to the gravitational force is

\frac{F_E}{F_G}=\frac{2.56\cdot 10^{-19} N}{1.13\cdot 10^{-68}N}=2.27\cdot 10^{49}

So, we see that the electrical force is much larger than the gravitational force.

5 0
3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

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You can describe the motion of an object by its position, speed, direction, and acceleration
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