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3 years ago

I need help ;-;

2 answers:

8 0

The First Quarter Moon is a primary Moon phase when we can see exactly half of the Moon's visible surface illuminated. If it is the left or right half, depends on where you are on Earth.

koban [17]3 years ago

5 0

Moon is a primary moon phase.

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When using science to investigate physical phenomena, which of the following is NOT a characteristic of the event must exist?

Andreas93 [3]

It’s not Measurable.

6 0

3 years ago

Read 2 more answers

What potential difference is required across a 8Ω resistor to cause 2.0 A to flow through it?

dalvyx [7]

Answer:

16 V

Explanation:

7 0

3 years ago

A movie stunt performer is filming a scene where he swings across a river on a vine. The safety crew must use a vine with enough

Julli [10]

Answer:

1125.66956 N

Explanation:

m = Mass of stunt performer

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of the swing = 7 m/s

T = Tension

r = Radius of the swing = Length of vine = 11.5 m

From the free body diagram

$T-mg-m\frac{v^2}{r}=0\\\Rightarrow T=mg+m\frac{v^2}{r}\\\Rightarrow T=m(g+\frac{v^2}{r})\\\Rightarrow T=80(9.81+\frac{7^2}{11.5})\\\Rightarrow T=1125.66956\ N$

The minimum tension force the vine must be able to support without breaking is 1125.66956 N

7 0

3 years ago

Describe two ways to change the frictional force between two solid surfaces.

Tanya [424]

<span>There are several ways to change the frictional force between two objects. The first one is to modify the surfaces of each object that will come in contact with each other. The smoother they get, the less friction there will be. But if the surfaces become rougher, more friction will be generated. If you don’t want to alter the surfaces, you can simply add lubrication to reduce friction.</span>

5 0

4 years ago

In the Bohr model of the hydrogen atom, an electron({rm mass};m=9.1; times 10^{ - 31;}{rm kg}) orbits a proton at a distance of

max2010maxim [7]

Answer:

n=6.56×10¹⁵Hz

Explanation:

Given Data

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

To find

Revolutions per second

Solution

Let F be the force of attraction

let n be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by

F=mω²r......................where ω=2π n

F=m4π²n²r...............eq(i)

as the values given where

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

we have to find n from eq(i)

n²=F/(m4π²r)

$n^{2} =\frac{8.2*10^{-8} }{9.11*10^{-31}* 4\pi^{2} *5.3*10^{-11} }\\ n^{2}=4.31*10^{31}\\ n=\sqrt{4.31*10^{31}}\\ n=6.56*10^{15}Hz$

8 0

3 years ago