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2 years ago

I need the answer please help

Physics

2 answers:

Sergeeva-Olga [200]2 years ago

8 0

Answer:

Option A

Inertia = the resistance an object has to a change in its state of motion.

timurjin [86]2 years ago

3 0

Answer:

A) Inertia

Explanation:

Inertia is why objects tend to resist changes in their motion. Like a rolling ball will keep rolling unless we try to add friction to it, which then would stop the ball.

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A 3 mm inside diameter tube is placed in a fluid with a surface tension of 600 mN/m and density of 3.7 g/cm3. The contact angle

Aleks04 [339]

Answer: The height of the fluid rise is 0.01m

Explanation:

Using the equation

h = (2TcosѲ )/rpg

h= height of the fluid rise

diameter of the tube =3mm

radius of the tube= 3/2 =1.5mm=0.0015

T= surface tension = 600mN/m=0.6N/m

Ѳ = contact angle = $60^o$ C

p= density =3.7g/cm3= 3700kg/m3

g= acceleration due to gravity =9.8m/s2

h = ( 2*0.6*0.5)/(0.0015*3700*9.8)

h = 0.6/54.39

h= 0.01m

Therefore,the height of the fluid rise is 0.01m

8 0

3 years ago

I NEED HELP

Oduvanchick [21]

Answer:

the answer is different

Explanation:

i took the test

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

3 years ago

An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec

aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

$x = V*t = -8\frac{m}{s} *3s = -24m$

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

$v^2 = v_0^2 + 2a*d$

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

$d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m$

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

$t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s$

Then, the time of the second phase will be:

$t_2 = 9s - t_1 = 9s - 1.143s = 7.857s$

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

$x = \frac{1}{2}a*t^2 + v_0*t + x_0$

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

$x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m$

So, the total distance covered by this object in meters will be the sum of all the distances we found:

$x_total = 24m + 4.57m + 216.07m = 244.64m$

8 0

2 years ago

Is there a definite end to our atmosphere?

Irina18 [472]

There is no definite end to earths atmosphere, but technically the border between the outer space and earth gets thinner as you move up from the earths surface. The Karman line is the closest definition there is which describes the end of the earth's atmosphere, it is 100 km above earth's sea level at approximately 1.56 % of total earth's radius. This describes the boundary between the outer space and the atmosphere.

7 0

3 years ago