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3 years ago

10

Fill out the blanks, this is for science!

2 answers:

dusya [7]3 years ago

8 0

Faster and higher I believe.

rewona [7]3 years ago

4 0

Answer:

Faster and higher

Explanation:

You might be interested in

What is a possible explanation for a set of facts that can be tested by further investigation called

Amanda [17]

Answer:

A hypothesis

Explanation:

A hypothesis is an explanation for those facts and can expierimented on.

5 0

3 years ago

Read 2 more answers

A student uses a stopwatch to measure the period of the pendulum of the Beverly clock in the corridor. His measurements are: (a)

Westkost [7]

Answer:

Reading is close to (b) 13.44 which is the best estimate of the period

Associated error, $\Delta E =0.178 s$

Given:

$t_{a} = 13.54 s$

$t_{b} = 13.44 s$

$t_{c} = 13.89 s$

$t_{d} = 13.41 s$

$t_{e} = 13.17 s$

$t_{f} = 13.22 s$

Solution:

1.The best estimate of the period can be calculated by the mean of the measurements and the one closest to the mean is the best estimate of the measurement:

$Mean, \bar {x} = \fra{sum of all observations}{No. of observation}$

$Mean, \bar {x} = \frac{t_{a} + t_{b} + t_{c} +t_{d} + t_{e} + t_{f}}{6}$

$Mean, \bar {x} = \frac{13.54 + 13.44 + 13.89 + 13.41 + 13.17 + 13.22}{6}$

$Mean, \bar {x} = 13.445 s$

It is close to 13.44 s

2. Associated error is given by:

$\Delta E_{n} = |measured value - actual value|$

$\Delta E_{n} = |t_{n} - \bar {x}|$

where

n = a, b,......, e

Now,

$\Delta E_{a} = |t_{a} - \bar {x}| = |13.54 - 13.44| = 0.01$

$\Delta E_{b} = |t_{b} - \bar {x}| = |13.44 - 13.44| = 0.00$

$\Delta E_{c} = |t_{c} - \bar {x}| = |13.89 - 13.44| = 0.45$

$\Delta E_{d} = |t_{d} - \bar {x}| = |13.41 - 13.44| = 0.03$

$\Delta E_{e} = |t_{e} - \bar {x}| = |13.17 - 13.44| = 0.027$

$\Delta E_{f} = |t_{f} - \bar {x}| = |13.54 - 13.44| = 0.10$

Mean Absolute Error, $\Delta E = \frac{\Sigma E_{n}}{6}$

$\Delta E = \frac{0.01 + 0.00 + 0.45 + 0.03 +0.027 + 1.10}{6}$

$\Delta E =0.178 s$

3. The assumption behind the estimation is population is considered to distributed normally.

6 0

3 years ago

b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force

Oduvanchick [21]

Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

$= force \times radius$

$= f \times r$

And,

Torque is also

$= moment\ of\ inertia \times angular\ acceleration$

$= I \times \alpha$

So,

We can say that

$f \times r = I \times \alpha$

$f \times 0.05 = 0.2 \times 2$

0.05f = 0.4

f = 8 N

We simply applied the above formulas

8 0

4 years ago

Suppose the roller-coaster car in fig.6–41 passes point 1 with a speed of if the average force of friction is equal to 0.23 of i

madreJ [45]

The values in figure 1 shows h1 = 39 m, h2 = 13 m, h3 = 25 with a speed of 1.5m/s

The Initial speed u = 1.5m/s.

Vertical distance covered between point 1 and point 2 is (39-13) =26m

The expanse 45 m is equal to a height of (45/6) =7.5 (As frictional force is mg/6)

With the kinematics equation v^2 =u^2 + 2as

V^2 = 1.5^2 + 2*9.8 (26 + 7.5)

V = 25.7 m/s

8 0

4 years ago

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land

andrey2020 [161]

Answer:

a).

The magnitude force impulse is

$F=656.6 kg*m/s$

the average force is

$F=656.6 kg*m/s$

Explanation:

Using the conservation energy the potential energy in the high is equal to the kinetic energy just before the collision so:

$E_K=E_p$

$\frac{1}{2}*m*v^2=m*g*h$

Notice the mass can be cancel as a factor so:

$v^2=2*g*h$

$v=\sqrt{2*g*h}=\sqrt{2*9.8m/s^2*0.50m}$

$v=9.8 m/s$

Now the impulse is determinate by:

$F=m*(v_f-(-v_i))$

$kg*F=m*v_i=67*9.8m/s$

$F=656.6 kg*m/s$

The average force is also

$F'=\frac{F}{t}=\frac{656.6}{0.095s}$

$F'=6911.57 N$

8 0

4 years ago