All categories

3 years ago

A 0.59 kg bullfrog is sitting at rest on a level log. how large is the normal force of the log on the bullfrog?

1 answer:

7 0

<span>The bullfrog is sitting at rest on the log. The force of gravity pulls down on the bullfrog. We can find the weight of the bullfrog due to the force of gravity. weight = mg = (0.59 kg) x (9.80 m/s^2) weight = 5.782 N The bullfrog is pressing down on the log with a force of 5.782 newtons. Newton's third law tells us that the log must be pushing up on the bullfrog with a force of the same magnitude. Therefore, the normal force of the log on the bullfrog is 5.782 N</span>

You might be interested in

Two bullets of the same size, mass and horizontal velocity are fired at identical blocks, only one is made of steel and the othe

tatiyna

Answer and Explanation:

Since we're discussing shots, the significant thing is the way the energy is changed over as there is deceleration of the bullet to a halt when it hits something.

Kinetic Energy is relative to mass times speed squared, so in reality, the 2 cases given have practically indistinguishable Kinetic energy. The measure of energy is authoritative, so the two cases will do generally a similar harm given, obviously we look at situations when all the kinetic energy is spent.

One contrast that will be effectively obvious is that the weapon in the case of heavy bullet will recoil more.
One can consider energy assimilation as force times separation distance, and energy ingestion as a product of force and time.
Henceforth, the heavier yet more slow bullet with a similar energy will venture to every part of a similar separation in the engrossing material, but since of bigger force, will take a more drawn out time doing it.
It will along these lines, additionally, give a more noteworthy "kick" to the object that absorbs.

8 0

4 years ago

The acceleration due to gravity on jupiter is 2.5 times whats on earth. An object of mass 10kg is taken to Jupiter. What is the

Damm [24]

The mass is still 10 kg. But instead of weighing 98N as it does on Earth, it weighs 245N on Jupiter.

4 0

3 years ago

Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th

brilliants [131]

From the case we know that:

The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

#SPJ4

8 0

1 year ago

Choose the appropriate descriptor for the term cubic centimeter .

Arada [10]

Answer:volume

Explanation:

4 0

4 years ago

Read 2 more answers

A boat has a porthole window, of

jekas [21]

Answer:

F = 534.6[N]

Explanation:

We must find the pressure exerted by the water at the depth of the boat, by means of the following equation.

$P=Ro*g*h$

where:

Ro = density of sea water = 1027 [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = wáter Depth = 6.25 [m]

Now replacing:

$P=1027*9.81*6.25\\P=62967.93[Pa]$

The net force is:

$F = P*A\\F = 62967.93*0.00849\\F = 534.6[N]$

7 0

3 years ago