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2 years ago

A 0.54 kg bullfrog is sitting at rest on a level log. How large is the normal force of the log on the bullfrog?

Physics

1 answer:

RUDIKE [14]2 years ago

7 0

Answer:

Explanation:

Step one:

given data

mass of bullfrog= 0.54kg

Step two:

Required

The force F in Newton

From newton first law

F=mg

The acceleration due to gravity acting on the frog is 9.81m/s^2

hence

F=0.54*9.81

F=5.3N

The normal force is 5.3N

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Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

4 0

2 years ago

When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of

pentagon [3]

Answer:

7772.72N

Explanation:

When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.

Now which direction is the static friction, assume that it is pointing inward so

Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N

Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.

7 0

2 years ago

The most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-

just olya [345]

Answer:

1.99 parsecs.

Explanation:

We have been given that the most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years.

We know that one light year equals to 0.306601 parsecs. To convert 6.5 light-years to parsecs, we will multiply 0.306601 by 6.5.

$6.5\text{ Light-years}=6.5\times 0.306601\text{ Parsecs}$

$6.5\text{ Light-years}=1.9929065\text{ Parsecs}$

$6.5\text{ Light-years}\approx 1.99\text{ Parsecs}$

Therefore, Luhman 16 is approximately 1.99 parsecs away from the Earth.

5 0

2 years ago

A uniform solid sphere has mass m= 7 kg and radius r= 0. 4 m. What is its moment of inertia about an axis tangent to its surface

lilavasa [31]

The moment of inertia of a uniform solid sphere is equal to 0.448 $kgm^2$ .

<u>Given the following data:</u>

Mass of sphere = 7 kg.

Radius of sphere = 0.4 meter.

<h3>How to calculate moment of inertia.</h3>

Mathematically, the moment of inertia of a solid sphere is given by this formula:

$I=\frac{2}{5} mr^2$

<u>Where:</u>

I is the moment of inertia.

m is the mass.

r is the radius.

Substituting the given parameters into the formula, we have;

$I=\frac{2}{5} \times 7 \times 0.4^2\\\\I=2.8 \times 0.16$

I = 0.448 $kgm^2$ .

Read more on inertia here: brainly.com/question/3406242

4 0

1 year ago

Convert 56 kilometers to inches

svlad2 [7]

The conversion for km to inches is:

1km=39370.1in

Now we can solve for 56 km..

56km=39370.1*56
56km=<span> 2204725.6in

Answer=2,204,725.6in</span>

7 0

2 years ago