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Romashka-Z-Leto [24]
3 years ago
7

A 10 kg boat is moving 3 m/s. Find kinetic energy.

Physics
2 answers:
mr Goodwill [35]3 years ago
7 0
KE = 1/2mv^2
m=10 v=3
KE=1/2*10*3^2
KE= 45 Joules
REY [17]3 years ago
4 0

answer

45J

explanation

kinetic energy can be found through the equation K = \frac{mv^2}{2}

plug in 10 for m (mass) and 3 for v (velocity)

K = \frac{mv^2}{2}\\K = \frac{10*3^2}{2}\\K = \frac{10*9}{2}\\K = \frac{90}{2}\\K = 45 J

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You hold a spherical salad bowl 50 cm in front of your face with the bottom of the bowl facing you. The salad bowl is made of po
nataly862011 [7]

Answer:

a) q = 39.29 cm ,  b)   h ’= - 3.929 cm  the image is inverted  and REAL

Explanation:

For this exercise we will use the equation of the constructor

          1 / f = 1 / p + 1 / q

where f is the focal length of the salad bowl, p and q are the distance to the object and the image

The metal salad bowl behaves like a mirror, so its focal length is

           f = R / 2

           f = 44/2

           f = 22 cm

a) Suppose that the distance to the object is p = 50 cm, let's find the distance to the image

           1 / q = 1 / f  - 1 / p

           1 / q = 1/22 - 1/50

           1 / q = 0.0254

            q = 39.29 cm

b) to calculate the size of the image we use the equation of magnification

           m = h’/ h = - q / p

            h ’= - q / p h

            h ’= - 39.29 / 50 5

            h ’= - 3.929 cm

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4 0
4 years ago
A missle is fired horizontally with an initial velocity of 45 m/s from the top of a building 75 m high.
NARA [144]

The horizontal range of the missile is b) 176 m

Explanation:

The motion of the missile is a projectile motion, so it consists of two independent motions:  

- A uniform motion with constant velocity along the horizontal direction  

- A uniformly accelerated motion with constant acceleration (equal to the acceleration of gravity) in the vertical-downward direction  

To find the time of flight of the missile, we study the vertical motion. We can use the following suvat equation:

s=u_y t+\frac{1}{2}at^2

where:

s = 75 m is the vertical displacement of the missile (the height of the building)

u_y=0 is the initial vertical velocity  (the missile is thrown horizontally)

t is the time of flight

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t, we find the time of flight:

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(75)}{9.8}}=3.91 s

This means that the missile takes 3.91 s to reach the ground.

Now we study the horizontal motion: the missile moves with a constant horizontal velocity of

v_x = 45 m/s

Therefore, the distance covered in a time t is

d=v_x t

and by substituting t = 3.91 s, we find the horizontal range of the missile:

d=(45)(3.91)=176 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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