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dimaraw [331]
4 years ago
5

PLEASE HELP AND HURRY

Physics
1 answer:
andreyandreev [35.5K]4 years ago
7 0
C
I took the tests earlier hope this helps
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If a car accelerates from rest at a constant 4 m/s
Readme [11.4K]

Answer:

The time it will take for the car to reach a velocity of 28 m/s is 7 seconds

Explanation:

The parameters of the car are;

The acceleration of the car, a = 4 m/s²

The final velocity of the car, v = 28 m/s

The initial velocity of the car, u = 0 m/s (The car starts from rest)

The kinematic equation that can be used for finding (the time) how long it will take for the car to reach a velocity of 28 m/s is given as follows;

v = u + a·t

Where;

v = The final velocity of the car, v = 28 m/s

u = The initial velocity of the car = 0 m/s

a = The acceleration of the car = 4 m/s²

t = =The time it will take for the car to reach a velocity of 28 m/s

Therefore, we get;

t = (v - u)/a

t = (28 m/s - 0 m/s)/(4 m/s²) = 7 s

The time it will take for the car to reach a velocity of 28 m/s, t = 7 seconds.

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3 years ago
Need help!!! A charge of 8.5 × 10–6 C is in an electric field that has a strength of 3.2 × 105 N/C. What is the electric force a
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Newton’s third law in football?
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Newtons third law of motion applies to football as when the ball is thrown, force is applied to the football thrown forward and also force back to the player. Therefore it states that every action has a reaction. The player does not get a huge force back from throwing the football due to the player having a bigger mass than the football being thrown.

In short, when the football is throw, there is a reaction with the ball and the player accordingly with the force put into the ball.
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What do the spheres in this model represent?
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A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the
Elena L [17]

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

8 0
3 years ago
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