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3 years ago

The center of mass is

1 answer:

5 0

D is the best answer. In many physics problems we treat an extended object as if it were a point with the same mass located at the center of mass.

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Describe the behavior of magnets

tatyana61 [14]

They have a north and South Pole .The opposite poles attract to each other ,the same poles repel from each other.

8 0

3 years ago

A 50 W light bulb is plugged into a standard

wolverine [178]

Answer:

$1.26

Explanation:

Power =energy/ time

energy =powerxtime

energy =50x31x24=37200

=37.2kwh

1kwh =3.39

37.2kwh=3.39x37.2=126.108cent

=$1.26

8 0

2 years ago

Jane is sliding down a slide. What kind of motion is she demonstrating? A. translational motion B. rotational motion C. vibratio

Andreyy89

A. translational motion

5 0

3 years ago

An object is 39 cm away from a concave mirrors surface along the principles axis. If the mirrors focal length is 9.50 cm, how fa

Tatiana [17]

Answer:

12.6 cm

Explanation:

We can use the mirror equation to find the distance of the image from the mirror:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where here we have

f = 9.50 cm is the focal length

p = 39 cm is the distance of the object from the mirror

Solving the equation for q, we find:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{9.50 cm}-\frac{1}{39 cm}=0.080 cm^{-1}\\q = \frac{1}{0.080 cm}=12.6 cm$

5 0

3 years ago

A burglar attempts to drag a 108 kg metal safe across a polished wood floor Assume that the coefficient of static friction is 0.

V125BC [204]

Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

Pushing force applied by the burglar, F = 534 N

Co-efficient of kinetic friction, $\mu_k$ = 0.3

Now,

The force against the kinetic friction is given as:

$f = \mu_k N = u_k Mg$

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

$f = 0.3\times108\times9.8$

$f = 317.52N$

Now, the net force on to the metal safe is

$F_{Net}= F-f$

Substituting the values in the equation we get

$F_{Net}= 534N-317.52N$

$F_{Net}= 216.48$

also,

$F_{Net}= M\times$ acceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe= $\frac{F_{Net}}{M}$

acceleration of the safe= $\frac{216.48}{108}$

acceleration of the safe= $2.00 m/s^2$

Hence, the acceleration of the metal safe will be 2.00 m/s²

3 0

3 years ago