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Damm [24]
3 years ago
14

5. A car has a kinetic energy of 4.32x105 J when traveling at a speed of 23 m/s. What is

Physics
1 answer:
DIA [1.3K]3 years ago
7 0

Answer:

m = 163.33 kg

Explanation:

KE = 4.32x10^5

V = 23

KE = 1/2 mv^2

4.32x10^5 = 1/2 * m * 23^2

8.64x10^5 = 529m

m = 163.33 kg

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posledela
The correct answer is this one: "D) significantly more than 100 billion gallons ." Clouds dump around 100 billion gallons of water on rainforests each year. The amount of  rain is evaporated from the rivers, lakes and surface of rainforests each year is significantly more than 100 billion gallons<span> </span>
3 0
3 years ago
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A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes
Alchen [17]
<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

x=V_{o}cos\theta t   (1)

Where:

V_{o}=54.5m/s is the projectile's initial speed

\theta=35\° is the angle

t=2.80s is the time since the projectile is launched until it strikes the target

x  is the final horizontal position of the projectile (the value we want to find)

y-component:

y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}   (2)

Where:

y_{o}=0  is the initial height of the projectile (we are told it  was launched at ground level)

y  is the final height of the projectile (the value we want to find)

g=9.8m/s^{2}  is the acceleration due gravity

Having this clear, let's begin with x (1):

x=(54.5m/s)cos(35\°)(2.8s)   (3)

x=125m   (4)  This is the horizontal final position of the projectile

For y (2):

y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}   (5)

y=48.308m   (6)  This is the vertical final position of the projectile

4 0
3 years ago
un movil que parte del reposo alcanza una velocidad de 75 m/s en 13 segundos ¿cual su aceleracion y el espacio que recorrio en l
Dmitriy789 [7]

Answer:

Acceleration = 5.77 m/s²

Distance cover in 13 seconds = 487.56 meter

Explanation:

Given:

Final velocity of mobile device = 75 m/s

initial velocity of mobile device = 0 m/s

Time taken = 13 seconds

Find:

Acceleration

Distance cover in 13 seconds

Computation:

v = u + at

75 = 0 + (a)(13)

13a = 75

a = 5.77

Acceleration = 5.77 m/s²

s = ut + (1/2)(a)(t²)

s = (0)(t) + (1/2)(5.77)(13²)

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8 0
2 years ago
A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incl
sp2606 [1]

Answer:

7.9m/s

Explanation:

We are given that

Mass of wagon=40 kg

\theta=18.5^{\circ}

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a=\frac{F}{a}=\frac{15.7}{40}=0.39m/s^2

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v^2-u^2=2as

Using the formula

v^2=2\times 0.39\times 80

v=\sqrt{2\times 0.39\times 80}=7.9m/s

5 0
3 years ago
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