A dog has a weight of 700N find the mass of the dog

What is the least force you could exert on the refrigerator to move it?

Ulleksa [173]

If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365

Fsf = Static frictional force = (coefficient of static friction) * (Normal force)

So the least for you could exert to move it is equal to the Fsf.
Fsf = (0.49)(1365N)

8 0

3 years ago

Un cuerpo de masa m tiene una velocidad v. si la masa del cuerpo se incrementa al triple de su valor original y la velocidad se

serg [7]

Urrrrrrrr nope now clue at all

5 0

3 years ago

How much would a 75kg man weigh due to the effect of gravity

Nezavi [6.7K]

Answer: 735 N

Explanation:

Weight $W$ is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass $m$ of the body by the acceleration of gravity $g$ :

$W=m.g$

In the case of our planet Earth, the acceleration due gravity is $g=9.8 m/s^{2}$ . So for a man whose mass is $m=75 kg$ , his weight is:

$W=(75 kg)(9.8 m/s^{2})$

$W=735 N$

4 0

3 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago

From an h = 53 feet observation tower on the coast, a Coast Guard officer sights a boat in difficulty. The angle of depression o

maksim [4K]

Answer:

757,93 feets

Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

The angle of depression, its O in the graph.

Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or legs).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

$tangent( O ) = \frac{opposite}{adjacent}$

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

$tangent( 4) = \frac{53 feets}{DB}$

$DB= \frac{53 feets}{tangent( 4)}$

$DB= 757,93 feets$

This its equal to the distance from the boat to the shoreline.

4 0

3 years ago