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galina1969 [7]
3 years ago
8

If you pull up on a bucket with a tension force of 15 N and the bucket has a weight of 15 N, what is the net force acting on the

bucket?
15 Nup
15 N down
30 N up
O N
Physics
1 answer:
fredd [130]3 years ago
3 0

Answer:

F = 0 [N]

Explanation:

To solve this problem we must perform a summation of forces in the direction of the vertical axis. Where the positive force is that of the tension of the upward force, while the force exerted by the weight is directed downward with a negative sign.

ΣF = 0

15 - 15 + F = 0

F = 0 [N]

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WILL MARK BRAINLIEST Why do we see sedimentary rocks more often than igneous and metamorphic rocks?
Arisa [49]

Answer:

because they are the rocks that line the surface of our planet ​

Explanation:

We see sedimentary rocks more than other rock types because they are the rocks that line the surface of our planet.

Sedimentary rocks typically form the earth cover due to the way they are formed.

  • These rocks are produced by the weathering, transportation and deposition of sediments within a basin.
  • In this basin, the sediment is lithified and converted to sedimentary rocks.
  • These processes are driven by the external heat engine
  • Therefore, it is confined to the surface.
  • Igneous and metamorphic rock's processes are confined to the subsurface.
8 0
3 years ago
A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?
anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\  \end{gathered}

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}

now, replace

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}

hence, the answer is 0.495 m/s

3 0
1 year ago
-Which of the following wires will have the least resistance?
scoray [572]
The wires that will have the least resistance is :
C. A short thick wire
in order to get the least resistence, you need the wire to be the lowest in length and the highest in Area

hope this helps
7 0
3 years ago
Read 2 more answers
The Bugatti Chiron is currently the fastest production car in the world reaching a top speed of 261 mph
VMariaS [17]

Answer:304 not 261

Explanation:

That is, in prototype form anyway – the custom Chiron Longtail hit 304 mph (490 km/h) last year, but the production version dubbed the Chiron Super Sport 300+ will be electronically limited to only 273 mph (440 km/h).

...

7 0
3 years ago
Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each mo
postnew [5]

Answer:

ucosφ=-v2cosθ2\\\\φ=cos^{-1} (\frac{-v2cosθ2}{cosφ} )

Explanation:

Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each moving at the same speed v when they collide at the origin of the coordinates and stick together. After the collision, the masses move at an angle −θ2 with respect to the +x axis at speed v2 .1. What was the angle φ?

from the principle of momentum

In a system of colliding bodies,we know that the total momentum before collision will equal to the total momentum after collision.

Take note that momentum is the product of mass and velocity

momentum before collision=momentum after collision

mass, m

u=initial velocity of the identical masses

v2=the common velocity after the collision

Note that the collision is inelastic , since they both moved with the same velocity

umcosφ+umcosφ=(m+m)v2cos−θ2

2mucosφ=2mv2cos−θ2

ucosφ=-v2cosθ2\\\\φ=cos^{-1} (\frac{-v2cosθ2}{cosφ} )

8 0
3 years ago
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