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Lubov Fominskaja [6]
3 years ago
10

A rectangular plate has a length of (22.4 ± 0.2) cm and a width of (8.4 ± 0.1) cm. Calculate the area of the plate, including it

s uncertainty.
Physics
1 answer:
blsea [12.9K]3 years ago
8 0
Area of the rectangle = Length  x  Width
We will calculate the percentage of uncertainties. 
For a length: p 1 = ( 0.2 · 100 ) : 22.4 = 20 : 22.4 = 0.893
For a width: p 2 =  ( 0.1  · 100 ) : 8.4 = 10 : 8.4 = 1.190
p 1 + p 2 = 0.893 + 1.190 = 2.083 %
A = 22.4 · 8.4 = 188.16 +/-  2.083 %
 Uncertainty: 188.16 · 0.02083  = 3.92 
A = 188.16 +/- 3.92 cm²     
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A student obtained a clean dry glass-stoppered flask. she weighed the flask and stopper on an analytical balance and found the t
svetoff [14.1K]
29.213 cm3


First, calculate the mass of the water used. You do this by subtracting the original mass of the flask from the combined mass of the water and flask, giving:60.735 g - 31.601 g = 29.134 g
So we now know we have 29.134 g of water. To calculate the volume of the flask, simply divide by the density of the water, giving:29.134 g / (0.9973 g/cm3) = 29.213 cm3
6 0
3 years ago
An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/
Katen [24]
<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}    (1)

Where:

y  is the instrument's final position  

y_{o}=0  is the instrument's initial position

V_{o}=15m/s is the instrument's initial velocity

t is the time the parabolic movement lasts

g=2.5\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of planet X.

As we know y_{o}=0  and y=0 when the object hits the ground, equation (1) is rewritten as:

0=V_{o}.t-\frac{1}{2}g.t^{2}    (2)

Finding t:

0=t(V_{o}-\frac{1}{2}g.t^{2})   (3)

t=\frac{2V_{o}}{g}   (4)

t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}   (5)

Finally:

t=12s

3 0
3 years ago
A rocket burns fuel to create hot gases that explode violently out of the rocket engine. This explosion creates thrust. Thrust i
Lerok [7]

Answer:

Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

Explanation:

From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

5 0
3 years ago
5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on
kari74 [83]

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

(0.1)(v1) + 0 = 30.6

(0.1)(v1) = 30.6

v1 = 306 m/s

8 0
3 years ago
4. What is the instantaneous acceleration at t= 10 s?
vodka [1.7K]

Answer:

I am fairly certain the answer is 2m/s^2

Explanation:

6 0
3 years ago
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