Question 18 of 20

evaluate the force of a skateboard exerts on you if your mass is 60kg and you push on the skateboard with a force of 60N

Sav [38]

Every action has an equal or opposite reaction.
You weigh 60kg
So your acceleration is 6N / 60 kg = 0.1m/s^2

6 0

3 years ago

In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

hammer [34]

Answer:

1201 lbs

Explanation:

Given that in mammals, the weight of the heart is approximately 0.5% of the total body weight.

Let the weight of the heart of a mammal be H

And the weight of the total body be B

The linear model that can gives the heart weight in terms of the total body weight will be:

H = 0.005B

B.) To find the weight of the heart of a whale whose weight is 2.402 × 105 lbs, substitute the whole weight in the formula.

H = 0.005 × 2.402 × 10^5

H = 1201 lbs

Therefore, the weight of the heart of the whale is 1201 lbs

8 0

3 years ago

What is the mass of an object that has a weight of 110N ?

fgiga [73]

Weight (W) = 110 N
Acceleration due to gravity (g) = 9.8 m/s^2
Let the mass of the object be m.
By using the formula, W = mg, we get,
110 N = 9.8 m/s^2 × m
or, m = 110 N ÷ 9.8 m/s^2
or, m = 11.2 Kg

Answer:

The mass of the object is 11.2 Kg.

Hope you could get an idea from here.

Doubt clarification - use comment section.

3 0

3 years ago

An engine has an energy input of 125j and 35 j of that energy is transformed into useful energy

Tatiana [17]

The engine's efficiency is (35J)/125J) = 28% .

Do you have a different question to ask ?

5 0

3 years ago

A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At

NikAS [45]

To solve this problem it is necessary to apply the concepts related to the magnetic field.

According to the information, the magnetic field INSIDE the plates is,

$B=\frac{1}{2} \mu \epsilon_0 r$

Where,

$\mu =$ Permeability constant

$\epsilon_0 =$ Electromotive force

r = Radius

From this deduction we can verify that the distance is proportional to the field

$B \propto r$

Then the distance relationship would be given by

$\frac{r}{R} = \frac{B}{B_{max}}$

$r =\frac{B}{B_{max}} R$

$r = \frac{0.5B_{max}}{B_{max}}R$

$r = 0.5R$

On the outside, however, it is defined by

$B = \frac{\mu_0 i_d}{2\pi r}$

Here the magnetic field is inversely proportional to the distance, that is

$B \not\propto r$

Then,

$\frac{r}{R} = \frac{B_{max}{B}}$

$r = \frac{B_{max}{B}}R$

$r = \frac{B_{max}{0.5B_{max}}}R$

$r = 2R$

7 0

3 years ago