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Genrish500 [490]
3 years ago
8

What is hypothesis testing

Physics
1 answer:
Stells [14]3 years ago
5 0
Hypothesis testing is basically testing the results of a experiment to see weather your results are valid or not.
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a car goes from a velocity zero to a velocity of 15 meters per second East in 2.1 seconds. What is the car's acceleration?
leva [86]
31.5 would be the acceleration.
3 0
3 years ago
A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. Wh
mixas84 [53]

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

diameter of the ceiling fan, d = 90 cm = 0.9 m

angular speed of the fan, ω = 64 rpm

time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

d = vt = \omega r\times  t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

4 0
3 years ago
How does a pin hole camera work? And how do we see through them?
DIA [1.3K]
A pinhole camera<span> is a simple </span>camera<span> without a lens but with a tiny aperture, a</span>pinhole<span> – effectively a light-proof box with a small hole in one side. Light from a scene passes through the aperture and projects an inverted image on the opposite side of the box, which is known as the </span>camera<span> obscura effect.</span>
6 0
3 years ago
Read 2 more answers
a 2.35 water bucket is swung in a full cirlce of radius 0.824 m just fast enough so that the water doesn't fall out the top mean
tiny-mole [99]

Answer:

2.84 m/s

Explanation:

At the top position of the circular trajectory, the normal reaction is zero:

N = 0

So it means that the only force that is providing the centripetal force is the gravitational force (the weight of the bucket). Therefore we have:

mg = m \frac{v^2}{r}

where

m is the mass of the water bucket

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

v is the speed of the bucket

r = 0.824 m is the radius of the circle

Solving for v,

v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(0.824 m)}=2.84 m/s

4 0
3 years ago
A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th
blagie [28]

Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

t = 3.19 s

Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

v0x = 28.2 m/s

8 0
3 years ago
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