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3 years ago

What energy does a galloping horse and a speeding bullet represent

Physics

2 answers:

Alex73 [517]3 years ago

8 0

Since both of those things are <u>moving</u>,
they both have <em>Kinetic Energy</em>.

Darya [45]3 years ago

6 0

These are both forms of kinetic energy.

(: I hope this helps! :)

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Which astronomer gave his name to a telescope launched into space in 1990?

Mnenie [13.5K]

Answer:

Edwin Hubble

Explanation:

5 0

3 years ago

A camera lens with focal length f = 50 mm and maximum aperture f>2

Brut [27]

Answer:

The minimum distance between two points on the object that are barely resolved is 0.26 mm

The corresponding distance between the image points = 0.0015 m

Explanation:

Given

focal length f = 50 mm and maximum aperture f>2

s = 9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D

Substituting the given values, we get –

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5

Y’ = 0.0015 m

8 0

3 years ago

A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1

Bumek [7]

Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

4 0

2 years ago

ear the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of

sesenic [268]

Answer:

a) $V_{2/1}=0.8m/s$

b) The second runner will win

c) d = 10.54m

Explanation:

For part (a):

$V_{2/1} = V_{2} - V_{1} = 0.8m/s$

For part (b) we will calculate the amount of time that takes both runners to cross the finish line:

$t_{1} = \frac{X_{1}}{V_{1}}=\frac{250}{3.45}=72.46s$

$t_{2} = \frac{X_{2}}{V_{2}}=\frac{250+45}{4.25}=69.41s$

Since it takes less time to the second runner to cross the finish line, we can say the she won the race.

For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:

X1 = V1*t2 = 239.46m Since the finish line was 250m away:

d = 250m - 239.46m = 10.54m

6 0

3 years ago

Read 2 more answers

A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing a

hodyreva [135]

Answer:

5 m/s2

Explanation:

The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.

The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle

$a_c = \frac{v^2}{r} = \frac{30^2}{300} = \frac{900}{300} = 3 m/s^2$

So the magnitude of the total acceleration is

$a = \sqrt{a_c^2 + a_l^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 m/s^2$

4 0

3 years ago