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fiasKO [112]
3 years ago
11

A student wish to measure the gravitational acceleration g. She does it by releasing a small lead ball from rest and measures th

e time t it takes the ball to drop a distance s. g can then be found using the relation t2. She measures t using an automated digital timer which manual states that it measures drop times up to 500 ms with an uncertainty of 3 ms. She sets the system up such that s 100.0 t 0.3 mm (measured using a measurement tape), and the timer reads that the drop time is 144 1 mS. . Calculate her best estimate of g and its uncertainty . Discuss how she could set up her experiment such that she would achieve a better precision in a single drop, without needing a more precise timer or being able to measure distances with higher precision
Physics
1 answer:
Goryan [66]3 years ago
5 0

Answer:

(9.64 +- 0.86) m/s^2

Explanation:

The generic motion equation for constant acceleration is

x = X0 + v0 * t + \frac{1}{2}*a * t^2

Where

X0: initial position

v0: initial speed

a: acceleration

t: time

If the object has an initial speed of zero, and the frame of reference is set conveniently so that the object initial position is zero, the equation simplifies to:

x = \frac{1}{2}*a * t^2

And the acceleration can be obtained as:

a = 2*\frac{x}{t^2}

Where x is the distance fallen and a = g.

So, with the data x = (100.0 +- 0.03) mm and t = (144 +- 3) ms we can calculate

g = 2*\frac{100}{144^2} = 9.64e-3 \frac{mm}{ms^2} = 9.64 \frac{m}{s^2}

For the uncertainty we have to calculate the relative uncertainties first

For the distance (100 * 0.3)/100 = 0.3%

For the time (100 * 3)/144 = 2.08%

For multiplications or divisions the relative uncertainties are added

0.3% + 2.08% + 2.08% = 4.46%

We convert this into absolute uncertainty:

(9.64e-3 * 4.46)/100 = 0.00043 mm/(ms^2)

Finally, this is multiplied by a constant scalar, so:

2 * 0.00043 mm/(ms^2) = 0.00086 mm/(ms^2)

We convert the units

0.86 m/(s^2)

And the measurement is (9.64 +- 0.86) m/s^2

A better method is putting the ball in a ramp instead of a free fall, that way the fall is longer and the effect of time measuring uncertainty is reduced.

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A satellite is in a circular orbit 8200 km above the Earth’s surface; i.e., it moves on a circular path under the influence of n
andreyandreev [35.5K]

Answer:

5.23km/s

Explanation:

Given

Radius of Earth = 6.37 * 10^6 m

Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m

Gravity Acceleration on Satellite Altitude = 1.87965m/s²

For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.

Centripetal Acceleration = V²/R

So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R

Make V the subject of formula

A = V²/R

V² = AR

V = √AR

Where R = (radius of earth) + (altitude of satellite)

R = 6.37 * 10^6 + 8.2 * 10^6

R = 14.57 * 10^6m

A = 1.87965m/s²

V = √(1.87965 * 14.57x10^6)

V = √27386500.5

V = 5233.211299001789

V = 5233.2113 m/s ------- Approximated

V = 5.23km/s

7 0
3 years ago
Rays traveling parallel to the principle axis of a concave mirror will reflect out through the mirrors focus?
abruzzese [7]

Answer:

True

Explanation:

When a ray travelling parallel to the principle axis of a concave mirror then the light ray reflect out through the mirrors and passing through the focus.

When a light ray travelling through focus of a concave mirror then after reflection the light ray reflect out through the mirror and  go parallel to principle axis.

Therefore, rays travelling parallel to the principle axis of a  concave mirror will reflect out through the mirrors focus.

It is true.

7 0
3 years ago
Read 2 more answers
Please help!!!!
Dafna11 [192]

The intensity of the electric field is 30,000 N/C

Explanation:

The strength of the electric field produced by a single-point charge is given by the equation

E=k\frac{q}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

q=3\cdot 10^{-9}C is the magnitude of the charge

r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity

Substituting, we find:

E=(8.99\cdot 10^9)\frac{3\cdot 10^{-9}}{(0.03)^2}=30,000 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
Which of the following statements are true?
inessss [21]

Answer:

a. If an object's speed is constant, then its acceleration must be zero.

FALSE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

so we can not say anything about the acceleration when speed is given to as and no information is given about velocity

b. If an object's acceleration is zero, then its speed must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

Since we know that if acceleration is 0 then velocity must be constant and hence speed is also constant

c. If an object's velocity is constant, then its speed must be constant.

TRUE

Since velocity is constant then it shows that its magnitude and direction both are constant so its speed is also constant.

d. If an object's acceleration is zero, its velocity must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

Since we know that if acceleration is 0 then velocity must be constant

e. If an object's speed is constant, then its velocity must be constant.

FALSE

Speed is just the magnitude so we can not say about its direction and hence if speed is constant then velocity may or may not change

7 0
3 years ago
Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

6 0
3 years ago
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