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3 years ago

What is the difference between applied force and normal force

Physics

2 answers:

Natalija [7]3 years ago

5 0

If someone is pushing a desk across the room then there is an applied force upon acting upon an object

jenyasd209 [6]3 years ago

3 0

If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person. The normal force is the support force exerted upon an object that is in contact with another stable object.

hope it helps

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Plyometric helps strengthen your bones true or false

MatroZZZ [7]

Plyometric does indeed help strengthen your bones

3 0

3 years ago

Giving a test to a group of students, the grades and gender are summarized belowGrades vs. Gender ABCMale10316Female465If one st

strojnjashka [21]

Answer:

20.45%

Explanation:

The probability that the student got a B is

$\frac{total\#of\text{ students that got B}}{\text{total students }}\times100\%$

Now, how many students are there in total?

The answer is

$10+3+16+4+6+5=44\; \text{students}$

How many students got a B?

The answer is

$3+6=9\; \text{students}$

therefore, the probability that the student has got a B is

$\frac{9\text{ students }}{44\text{ students }}\times100\%=20.45\%$

Hence, the probability that a student has got a B is 20.45%

5 0

2 years ago

What is the name of Newton's first law of motion?

Ne4ueva [31]

A) law of inertia. an object in motion stays in motion

3 0

3 years ago

A ball bearing of radius of 1.5 mm made of iron of density

Serjik [45]

Answer:

$\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}$

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine ( $\sf \eta$ )

Explanation:

$\boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}$

$\sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v}$

Substituting values of r, ρ, σ, v & g in the equation:

$\sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times 64.7196$

$\sf \implies \eta = 2 \times 7.191$

$\sf \implies \eta = 14.382 \: poise$

6 0

3 years ago

The Gulf Stream off the east coast of the United States can flow at a rapid 3.9 m/s to the north. A ship in this current has a c

Alex Ar [27]

Answer:

72.54 degree west of south

Explanation:

flow = 3.9 m/s north

speed = 11 m/s

to find out

point due west from the current position

solution

we know here water is flowing north and ship must go south at an equal rate so that the velocities cancel and the ship just goes west

so it become like triangle with 3.3 point down and the hypotenuse is 11

so by triangle

hypotenuse ×cos(angle) = adjacent side

11 ×cos(angle) = 3.3

cos(angle) = 0.3

angle = 72.54 degree west of south

3 0

3 years ago

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