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2 years ago

A horse runs 350 meters in 23 seconds. What was the average speed of the horse? Round your answer to a tenth

Physics

1 answer:

OLga [1]2 years ago

5 0

Answer:

Horse/Speed

55 mph

rounded to the tenth? either 60 or 50

but 350 would stay like that i believe!

Explanation:

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You are designing a new piece of fishing equipment that will allow you to catch a fish from the surface. You know that water ref

OleMash [197]

Answer:

Law of refraction

Explanation:

An experiment to analyze the refraction of light in water can easily be performed with a laser pointer and protractor.

We throw the fishing rod line into the water, place the protractor at the point where the line touches the water and use the direction of the line for the direction of the laser pointer (on), the laser is visible by the reflection on the particles in the air.

The vertical line is called Normal and all angles must be measured with respect to this reference in optics.

Having these angles and the refractive index of water we can use the law of refraction

n₁ sin θ₁ = n₂ sin θ₂

θ₂ = $sin^{-1} ( \frac{n_1}{n_2} \ sin \ \theta_1 )$

we can repeat several times to analyze several different input points (different angles) and to decrease the errors in the measurements.

the refractive index of air is n1 = 1 and n2= 1.33 (water)

4 0

2 years ago

A 0.43-kg object mass attached to a spring whose spring constant is 561 N/m executes simple harmonic motion. If its maximum spee

VLD [36.1K]

Answer:

0.22 m

Explanation:

m = 0.43 kg, K = 561 N/m

Vmax = 8 m/s

Let the amplitude of the oscillations be A.

The formula for the angular frequency of oscillation sis given by

$\omega = \sqrt{\frac{K}{m}}$

$\omega = \sqrt{\frac{561}{0.43}}$

ω = 36.1 rad/s

The formula for the maximum velocity is given by

Vmax = ω x A

A = Vmax / ω

A = 8 / 36.1 = 0.22 m

4 0

2 years ago

Question 5 of 10

ElenaW [278]

Answer:

B it is b because thermal will move from boiling water to the ice

4 0

3 years ago

A child on a sled has a combined mass of 54 kg. At the top of a 2.7 meter hill, the sled has a

leonid [27]

Answer:

8.78?m/s

Explanation:

There are many students who can not get answers step by step and on time

So there are a wats up group where you can get help step by step and well explained answer from trusted experts.

so hurry up

5 0

2 years ago

A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta

Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, $\theta = 54^{\circ}$

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, $R = \frac{v^{2}sin2\theta}{g}$

$R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m$

Time of flight, t' = $\frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s$

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

$s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m$

Distance traveled vertically:

$s_{y} = vcos\theta - \frac{1}{2}gt^{2}$

$s_{y} = 12.4sin54^{\circ}\times 1.42 - \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m$

Now,

$s_{x} = \frac{mx + m'x'}{M}$

$10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}$

x' = 3.535 m

Similarly,

$s_{y} = \frac{my + m'y'}{M}$

$4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m$

The location of the other fragment is at (3.535, 1.162)

5 0

3 years ago

A horse runs 350 meters in 23 seconds. What was the average speed of the horse? Round your answer to a tenth ​

A horse runs 350 meters in 23 seconds. What was the average speed of the horse? Round your answer to a tenth