PLEASE HELP ME GUYS IM DESPERATE!!!

Calculate the mass of -1.5C of electrons

algol13

The electron is a type of low-mass, very negatively charged with a particle. As such, it can easily be deflected by passing close to other electrons or the positive nucleus of an atom. m = mass of an electron in kg = 9.10938356 × 10-31 kilograms. e = magnitude of the charge of an electron in coulombs = 1.602 x 10-19 coulombs. Hope this helps!

8 0

3 years ago

An archer pulls the bowstring back for a distance of 0.470 m before releasing the arrow. The bow and string act like a spring wh

never [62]

Answer:

(a) 46.94 J.

(b) 55.95 m/s

Explanation:

(a)

Potential Energy: This is the energy of a body, due to its position. The S.I unit of potential energy is Joules (J).

The formula of potential energy in a stretched spring is

Ep = 1/2ke² .......................... Equation 1

Where Ep = potential energy of the spring, k = Force constant of the spring, e = extension or compression.

Given: k = 425 N/m, e = 0.47 m.

Substitute into equation 1

Ep = 1/2(425×0.47²)

Ep = 46.94 J.

(b)

at the instant When the arrow leaves the bow, the potential energy of the arrow is converted kinetic energy of the bow.

I.e,

Ep = 1/2mv² ............. Equation 2

Where m = mass of the arrow, v = velocity of the arrow.

make v the subject of the equation

v = √(2Ep/m)............. Equation 3

Given: Ep = 46.94 J, m = 0.03 Kg.

Substitute into equation 3

v = √(2×46.96/0.03)

v = √(93.92/0.03)

v = √(3130.67)

v = 55.95 m/s

5 0

2 years ago

Which statements accurately describe mechanical waves? Check all that apply.

kotegsom [21]

Answer:

A

Explanation:

E. An ocean wave moving through water is an example of a mechanical wave

e.g sound waves wave on a rope or string

and Ans a is also correct

4 0

2 years ago

What force is required to accelerate to 10 kg object to 5.9 m/s/s?

g100num [7]

Force required to accelerate 10 kg object to 5.9 m/s/s ?

Mass = 10 kg

Acceleration = 5.9 m/s^2

Force = Mass * Acceleration

Force = 10 kg * 5.9 m/s^2

Force = 59 kg m /s^2 = 59 N

3 0

2 years ago

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If

mart [117]

Answer: 29.50 m

Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

5 0

2 years ago