All categories

2 years ago

An 8 N force accelerates an object at a rate of 2.5 m/s what Is the mass of the object

Physics

2 answers:

shtirl [24]2 years ago

8 0

Answer: 3.2 kilograms

=================================================

Work Shown:

F = 8 newtons = force applied
m = unknown mass in kilograms
a = 2.5 m/s^2 = acceleration

Apply Newton's Second Law of Motion

F = m*a

8 = m*2.5

2.5m = 8

m = 8/2.5

m = 3.2

The mass is 3.2 kilograms

DaniilM [7]2 years ago

6 0

Answer:

16N

Explanation:

You might be interested in

a cyclist coasting down a 5.0 ◦ incline at a constant speed of 6.0 km/h because of air resistance. If the total mass of the bicy

Dvinal [7]

Answer:

$F_{net}= 85.41\ N$

Explanation:

mass of the bicycle + cyclist = 50 kg

constant speed = 6 km/h

a cyclist coasting down a 5.0° incline

the downward velocity is constant, so net acceleration must be zero

the air drag must be equal to gravitational force downward along the ramp

$F_a = mg sin \theta$

now for upward motion

$F_{net} = mg sin \theta + air\ drag$

$F_{net} = mg sin \theta + mg sin \theta$

$F_{net} = 2 mg sin \theta$

$F_{net} = 2\times 50 \times 9.8 sin 5^0$

$F_{net}= 85.41\ N$

3 0

3 years ago

A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

5 0

3 years ago

When a board with a box on it is slowly tilted to a larger and larger angle, common experience shows that the box will at some p

Shtirlitz [24]

Answer:

the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic frictionExplanation:

This exercise uses Newton's second law with the condition that the acceleration is zero, by the time the body begins to slide. At this point the balance of forces is

fr- w || = 0

The expression for friction force is that it is proportional to the coefficient of friction by normal.

fr = μ N

When the system is immobile, the coefficient of friction is called static coefficient and has a value, this is due to the union between the surface, when the movement begins some joints are broken giving rise to coefficient of kinetic friction less than static.

In consequence a lower friction force, which is why the system comes out of balance and begins to accelerate.

μ kinetic <μ static

In all this movement the normal with changed that the angle of the table remains fixed.

Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction

3 0

2 years ago

Why is it possible to see the moon during the day?

agasfer [191]

It depends on the brightness of its light—which is really just the sun's light reflecting off of the moon's<span> surface—and the earth's rotation. In order for the </span>moon<span> to be visible in the sky, it needs to be above the horizon. </span>

6 0

3 years ago

Is there any position in the elliptical orbit of a satellite where the tangential component of the acceleration is greater than

Wittaler [7]

your mom n:

is fjie nu.skvjs.,zndzfxoiv xjl,mdfk

7 0

2 years ago