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3 years ago

The relationship between body mass body weight and body density

Physics

1 answer:

nordsb [41]3 years ago

7 0

Answer:

The formula for calculating Density is:

= Mass / Volume

From this formula, we can say that the relationship between Mass and Density is a direct one. In other words, if mass is increasing - all else being equal - then density will increase as well.

If mass however was decreasing, density would have to decrease as well.

For example, assume 3 bricks have masses of 5kg, 10kg and 15kg. Also assume that the bricks all have the same volume of 5 m³.

Density of 5kg brick = 5 / 5 = 1 kg/m³

Density of 10kg brick = 10 / 5 = 2kg / m³

Density of 15kg brick = 15 / 5 = 3 kg /m³

<em>Notice how density increases as mass increases and decreases when mass decreases. </em>

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If you double the force an object what will hapen to the accelration

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Explanation:

the acceleration will double because force is directly proportional to the acceleration

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3 years ago

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Suppose that a car traveling to the west (-x direction) begins to slow down as it approaches a traffic light. Make a statement c

nata0808 [166]

Answer:

The car is decelerating and its acceleration is negative.

Explanation:

First thing to consider. The car traveling direction is not important to answer the question.

Second, there are two affirmations on each sentence. So, lets take definitions:

Deceleration: Reducing the acceleration, from one value to another. No matter if the acceleration for the moment is positive or negative.

Acceleration negative: the driver is applying a force to make speed=0, the sign of this force must negative in order for the car to stop, other way it will go faster. Acceleration is the change rate of speed, if the speed is going down, the total acceleration until cars stops must be negative.

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3 years ago

When you’re driving a golf ball, a good "follow-through" helps to increase the distance of the drive. A good follow-through mean

Natali [406]

Answer:

$v_{f}$ = F Δt / m

Explanation:

The impulse theorem and momentum is

I = Δp

F Δt = m $v_{f}$ - m v₀

Before touching the ball is without movement, the force is represented by the kinetic energy of the club when it reaches the bottom of the path, so if this force is constant, the time [or the duration of the blow is directly proportional to the speed that the ball acquires.

$v_{f}$ = F Δt / m

The higher this speed, the greater the distance traveled.

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4 years ago

A 15.7 kg block is dragged over a rough, horizontal surface by a constant force of 83.1 N

Kay [80]

Answer:

The work done by the 83.1 N force is 4687.5 J.

The magnitude of the work done by the force of friction is 1187.5 J.

Explanation:

Given:

Mass of the block, $m=15.7$ kg

Force acting on it, $F=83.1$ N

Angle of application of force, $\theta = 25.7$ °

Displacement of the block, $d=62.6$ m

Coefficient of friction, $\mu =0.161$

Acceleration due to gravity, $g=9.8$ m/s²

Work done by a force is given as:

$Work,W=F\times d\times \cos\theta$

So, work done by the constant force is given as:

$W_{force}=83.1\times 62.6\times \cos(25.7)\\W_{force}=4687.5\textrm{ J}$

Now, in order to find work done by friction, we need to evaluate friction.

For the vertical direction, the net force is zero as there is no vertical motion.

Therefore,

$N + F\sin\theta = mg\\ N = mg-F\sin\theta$

Frictional force is given as:

$f=\mu N=\mu (mg-F \sin \theta)=0.161\times ((15.7\times 9.8)-(83.1\times \sin(25.7))=18.97\textrm{ N}$

Now, friction acts in the direction opposite to the displacement. Thus, angle between frictional force and displacement is 180°.

Therefore, work done by friction is:

$W_{fric}=f\times d\times \cos\theta_f\\W_{fric}=12.72\times 62.6\times \cos(-180)\\W_{fric}=18.97\times 62.6\times -1\\W_{fric}=-1187.5\textrm{ J}$

Negative sign indicates that frictional force is acting opposite to motion.

So, the magnitude of the work done by the force of friction is 1187.5 J.

8 0

4 years ago

An archer shoots an arrow at a 74.0 m distant target, the bull's-eye of which is at same height as the release height of the arr

GuDViN [60]

A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °

B. The arrow will go over the branch.

<h3>A. How to determine the angle</h3>

Range (R) = 74 m
Initial velocity (u) = 33 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Angle (θ) = ?

R = u²Sine(2θ) / g

74 = 33² × Sine (2θ) / 9.8

Cross multiply

74 × 9.8 = 33² × Sine (2θ)

725.2 = 1098 × Sine (2θ)

Divide both sides by 1098

Sine (2θ) = 725.2 / 1098

Sine (2θ) = 0.6605

Take the inverse of sine

2θ = Sine⁻¹ 0.6605

2θ = 41.3

Divide both sides by 2

θ = 41.3 / 2

θ = 20.7 °

<h3>B. How to determine if the arrow will go over or under the branch</h3>

To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:

Initial velocity (u) = 33 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Angle (θ) = 20.7 °
Maximum height (H) = ?

H = u²Sine²θ / 2g

H = [33² × (Sine 20.7)²] / (2 ×9.8)

H = 6.94 m

Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).

Therefore, we can conclude that the arrow will go over the branch

Learn more about projectile motion:

brainly.com/question/20326485

#SPJ1

3 0

2 years ago