All categories

3 years ago

What occurs any time an object speeds up, slows down, or changes directions

1 answer:

4 0

Whenever the motion of an object changes . . . speeding up, or slowing down,
or changing direction . . . that change is called "acceleration". Acceleration is
produced by force on the object.

If there is no force on the object, then there is no acceleration. That means that
its motion doesn't change. The object remains in constant, uniform motion ...
moving with steady speed, in a straight line.

No force is necessary to keep an object moving, only to change its motion.

You might be interested in

One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached t

Jobisdone [24]

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2 $\frac{m}{s^{2} }$

Explanation:

First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:

$K_{1} +U_{1}+W_{ext}=K_{2}+U_{2}\\K_{1}=0\\U_{1}=\frac{1}{2} kx^{2} _{1} =0\\W_{ext}=F$ Δx= $(0.400m)\\$

W_{ext}=20.4 Nm

$U_{2} =\frac{1}{2} kx^{2} =\frac{1}{2} (76.0N/m)0.400^{2}=6.08Nm\\k_{2} =\frac{1}{2}mv^{2} _{2} \\\frac{1}{2} mv^{2} _{2}=W_{ext}-U_{2}\\v_{2}=\sqrt{\frac{W_{ext}-U_{2}}{m} } \\v_{2}=\sqrt{\frac{20.4Nm-6.08Nm}{2.5kg} } \\v_{2}=2.39 \frac{m}{s}$

To determine, the acceleration we solve the following equation for a:

$F=ma\\a=\frac{F}{m} =\frac{51.0N}{5.00kg}\\a=10.2\frac{m}{s^{2} }$

8 0

4 years ago

_____ provide a picture of an electric field indicating a field’s strength.Question 15 options:CapacitorsElectric potential diff

kaheart [24]

Electric field lines are a pictorial way to represent the electric field of a charge which basically indicates the field's strength.

The electric field is strong when the lines are close together and weak when they are far away from each other.

The greater the strength of the charge, the greater will be the number of field lines.

Hence, electric field lines provide a picture of an electric field indicating a field’s strength.

The correct option is Electric field lines

4 0

1 year ago

What type of wave is produced when the frequency of a visible light wave is increased

SIZIF [17.4K]

-- If the frequency of a wave is too high for our eyes to detect it ... but not too too high ... we call it an ultraviolet wave.

-- If the wave's frequency is even higher than that, we call it an X-ray wave.

-- If the wave's frequency is even higher than that, we call it a Gamma Ray wave.

7 0

4 years ago

How does a lever apply forces with advantage?

KATRIN_1 [288]

Answer:

quick maths

Explanation:

quick maths

3 0

3 years ago

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl

aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i: unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R= 131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

$D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2} }$

$D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2} }$

D₂= 167.21 cm

Direction of the second displacement

$\beta = tan^{-1} \frac{D_{y}}{D_{x} }$

$\beta = tan^{-1} \frac{62.18}{155.22 }$

β= 21.8°

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0

4 years ago