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Anna71 [15]
4 years ago
15

a 1600 kg car on flat ground is moving 6.25 m/s. its engine creates 1150 N forward force as the car moves 45.8 m. what is it fin

al KE?(unit=J)PLEASE HELP
Physics
1 answer:
Sveta_85 [38]4 years ago
6 0

Answer:

83,900 J

Explanation:

First, find the acceleration:

F = ma

1150 N = (1600 kg) a

a = 0.719 m/s²

Now find the final velocity.

Given:

Δx = 45.8 m

v₀ = 6.25 m/s

a = 0.719 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)

v = 10.2 m/s

Now find the final KE:

KE = ½ mv²

KE = ½ (1600 kg) (10.2 m/s)²

KE = 83,920 J

Rounded to three significant figures, the final kinetic energy is 83,900 J.

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Calculate the density of sulfuric acid is 35.4 mL of the acid weighs 65.14 g
anygoal [31]

Answer:

1.84 g/mL

Explanation:

Density = mass / volume

ρ = 65.14 g / 35.4 mL

ρ = 1.84 g/mL

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4 years ago
A parallel-plate capacitor with circular plates of radius 0.10 m is being discharged. A circular loop of radius 0.20 m is concen
Sedaia [141]

Answer:

7.1934 x 10^12 V/m.s

Explanation:

In order to do this exercise, you need to use the correct formula. Besides that, we need to identify our data.

First we have the radius of the plates which are circular, and it's 0.1 m. The current of the loop (I) is 2.0 A, and the radius of the loop is 0.2 m.

Now with this data, we use the next formula:

I = dE/dt Eo A

Where:

dE/dt = rate of electric field

Eo = constant of permittivity of free space

A = Area of circle

Solving for dE/dT:

dE/dt = I / Eo*A

Now, the area of the circle is A = πr²

A = 3.1416 * (0.1)² = 0.031416 m²

Now solving the electric field:

dE/dt = 2 / (8.85x10^-12 * 0.031416)

dE/dt = 7.1934 x 10^12 V/m.s

6 0
3 years ago
A behavior that is developed by observing others or being taught is
Hatshy [7]

B. Learned Behavior
5 0
3 years ago
Read 2 more answers
A piece of string 2 meters and has a mass of 5g. On one end of the string hangs a 200 g mass. Find the tension of the string and
viktelen [127]

Explanation:

It is given that,

Length of the string, l = 2 m

Mass of the string, m=5\ g=5\times 10^{-3}\ kg

Hanged mass in the string, m'=200\ g=0.2\ kg

1. The tension in the string is given by :

T=m'g

T=0.2\times 9.8

T = 1.96 N

2. Velocity of the transverse wave in the string is given by :

v=\sqrt{\dfrac{T}{M}}

m = M/l

v=\sqrt{\dfrac{Tl}{m}}

v=\sqrt{\dfrac{1.96\times 2}{5\times 10^{-3}}}

v = 28 m/s

Hence, this is the required solution.

4 0
3 years ago
A bin is given a push across a horizontal surface. The bin has a mass m, the push gives it an initial speed of 1.60 m/s, and the
emmasim [6.3K]

Answer:

The bin moves 0.87 m before it stops.

Explanation:

If we analyze the situation and apply the law of conservation of energy to this case, we get:

Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)

F d = (0.5)(m)(Vi² - Vf²)

where,

F = Frictional Force = μR    

but, R = Normal Reaction = Weight of Bin = mg

Therefore, F = μmg

Hence, the equation becomes:

μmg d = (0.5)(m)(Vi² - Vf²)

μg d = (0.5)(Vi² - Vf²)

d = (0.5)(Vi² - Vf²)/μg

where,

Vf = Final Velocity = 0 m/s (Since, bin finally stops)

Vi = Initial Velocity = 1.6 m/s

μ = coefficient of kinetic friction = 0.15

g = 9.8 m/s²

d = distance moved by bin before coming to stop = ?

Therefore,

d = (0.5)[(1.6 m/s)² - (0 m/s)²]/(0.15)(9.8 m/s²)

<u>d = 0.87 m</u>

5 0
3 years ago
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