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3 years ago

According to newton's first law, what is required to make an object slow down?

Physics

2 answers:

skad [1K]3 years ago

8 0

An equal and opposite force going against the object in motion. I hope this helps! :)

Luden [163]3 years ago

3 0

I believe it is friction

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Which of your groups should you NOT change anything for? (in other

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3 years ago

Read 2 more answers

When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of t

choli [55]

Answer:

Part a)

$\tau = 23.1 Nm$

Part b)

$\tau = 17.05 Foot pound force$

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

$\tau = \vec r \times \vec F$

now we have

$\tau = (0.140)(165)$

$\tau = 23.1 Nm$

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

$\tau = 23.1 Nm$

$\tau = 23.1 (0.225 Lb)(3.28 foot)$

$\tau = 17.05 Foot pound force$

3 0

3 years ago

A cat lies on the floor. can you say that no force acts on the cat? or is it correct to say that no net force acts on the cat? e

DIA [1.3K]

Odd though it seems at first, gravity is pulling the cat down while the floor is pushing the cat up - in equal amounts. Forces are absolutely acting on the cat but they balance - so there is no net force.

5 0

3 years ago

Weak magnetic fields can be measured at the surface of the brain. Although the currents causing these fields are quite complicat

STALIN [3.7K]

To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.

By definition we know that the magnetic field in a sphere can be described as

$B = \frac{\mu_0}{2}\frac{Ia^2}{(z^2+a^2)^{3/2}}$

Where,

a = Radius

z = Distance to the magnetic field

I = Current

$\mu_0 =$ Permeability constant in free space

Our values are given as

$D=2a = 16cm \rightarrow$ diameter of the sphere then,

$a = 0.08m$

Thus z = a

$B = \frac{\mu_0}{2}\frac{Ia^2}{(a^2+a^2)^{3/2}}$

$B = \frac{\mu_0I}{2(2^{3/2})a}$

$B = \frac{\mu_0 I}{2^{5/2}a}$

Re-arrange to find I,

$I = \frac{2^{5/2}Ba}{\mu_0}$

$I = \frac{2^{5/2}(3*10^{-12})(8*10^{-2})}{4\pi*10^{-7}}$

$I = 1.08*10^{-6}A$

Therefore the current at the pole of this sphere is $1.08*10^{-6}A$

5 0

2 years ago