All categories

3 years ago

8. What is the weight in newtons of a 10 kg mass on the earth's surface?

Physics

1 answer:

iogann1982 [59]3 years ago

7 0

Weight = Mass times Acceleration due to gravity.

Weight = 10 kg times 9.8 m/s^2

Weight = 98 kg

You might be interested in

How did the astronomy of hipparchus and ptolemy violate the principles of early greek philosophers?

Mariana [72]

Answer: Hipparchus & Ptolemy believed sun and moon travel around circles with Earth in the middle (geocentric) while Plato and Aristotle believed Perfect unchanging heavens

Explanation: thus hipparchus and ptolemy violate early greek philosophy

8 0

3 years ago

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

3 years ago

Help me I do not understand

ololo11 [35]

Answer:

170N

Explanation:

First add 530N to 150N and you get 680N, then add 400N to 450N and get 850N. So subtract 850N by 680N and you get 170N

4 0

3 years ago

Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the rig

riadik2000 [5.3K]

Answer:

Plss see attached file

Explanation:

3 0

3 years ago

A block with mass m is pulled horizontally with a force F_pull leading to an acceleration a along a rough, flat surface.

Simora [160]

Answer:

$\mu_k=\frac{a}{g}$

Explanation:

The force of kinetic friction on the block is defined as:

$F_k=\mu_kN$

Where $\mu_k$ is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:

$N=mg\\F_k=F=ma$

Replacing this in the first equation and solving for $\mu_k$ :

$ma=\mu_k(mg)\\\mu_k=\frac{a}{g}$

6 0

3 years ago