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xenn [34]
3 years ago
15

According to Newton’s Second Law of Motion, an object will accelerate if you apply what kind of force?

Physics
1 answer:
Alex777 [14]3 years ago
6 0

If you got this question from a book or homework sheet, then
you should take it back and comp[lain. It's a defective question.

It's fishing for the answer "unbalanced force", but that's a misleading
and misuse of the term.  There's no such thing as a "balanced force"
or an "unbalanced force". 

A GROUP of two or more forces can be balanced ... if the forces
all cancel each other out and add up to zero ... or unbalanced ...
if they don't.  But it's never correct to apply that description to
a single force.

According to Newton's second law of motion, an object will
accelerate if the forces on it are unbalanced.

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As the mass of an object increases, the weight of the object will ______?
Sergio039 [100]

Answer:

also increase

Explanation:

If the mass increases so will the weight.

4 0
2 years ago
The blades in a blender rotate at a rate of 7700 rpm. when the motor is turned off during operation, the blades slow to rest in
Tpy6a [65]

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (speed at the end) - (speed at the beginning)

For this fan, speed at the end = 7700 rpm, speed at the end = 0 .

Change in angular speed = -7700 rpm

Angular acceleration = (-7700 rpm) / (2.5 sec)

<em>Angular acceleration = -3,080 rev per minute / sec</em>

That's a perfectly good and true answer to the question, but the units are ugly.  We really need to fix the units, and convert them into something prettier before we hand in this assignment.

1 rev = 2π radians, and

1 minute = 60 seconds .

So

Angular acceleration =

(-3,080 rev/min-sec) · (2π rad/rev) · (1 min/60 sec)

AngAccel = (-3,080 · 2π · 1 / 60) · (rev·rad·min / min·sec·rev·sec)

AngAccel = ( -102 and 2/3 · π) · (rad/s²)

<em>AngAccel = -322.5 radian/s²</em>

7 0
3 years ago
3. A dog walks a distance of 100 ft. in 20 s of time. What is the dog's
natta225 [31]
The answer is A.
100/20=5
8 0
3 years ago
Read 2 more answers
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
Objects can be charged by the transfer of electrons.<br> True<br> False
Sveta_85 [38]

Answer:

True

Explanation:

Whenever electrons are transferred between objects, neutral matter becomes charged. For example, when atoms lose or gain electrons they become charged particles called ions. Three ways electrons can be transferred are conduction, friction, and polarization. In each case, the total charge remains the same.

I tried, hope this helps :)

* I might be wrong though

4 0
3 years ago
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