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Nostrana [21]
3 years ago
14

Mr. Nakamora wants to test the accuracy of an old balance in his classroom compared to a brand-new balance that he knows is accu

rate. He weighs an object on the old balance and finds it to have a mass of 15.6 grams. The same object weighs 15.9 grams on the brand-new balance
What is the percent error between the old and new balance?
Physics
2 answers:
Juli2301 [7.4K]3 years ago
8 0

the answer is d                     ..........................................

faust18 [17]3 years ago
7 0
It 5.009084564 and you probabably are correct

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A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple
aliina [53]

Answer:

T= 1 s

Explanation:

Given that

When x=  cm ,T= 1

we know that time period of spring mas system given as

T=2\pi \sqrt{\dfrac{m}{k}}

T= Time period

m= mass

k=spring constant

So from above equation we can say that time period of system does not depends on the value of x.

So when x= 10 cm ,still time period will be 1 s.

T= 1 s

5 0
3 years ago
A car has the velocity of 2.35 after 4.67 it's velocity is 9.89 what is the average acceleration
kiruha [24]
Average acceleration  =  (change in speed) / (time for the change) .
  
Change in speed = (ending speed) - (beginning speed)

                       =  (9.89 miles/hour) - (2.35 yards/second)  = 26,839.2 ft/hr

Acceleration  =  (26,839.2 ft/hr) / (4.67 days)  =  2,873.58 inch/hour²  
6 0
3 years ago
Help pls i need this right now
pantera1 [17]

Answer:

The x-component of F_{3} is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O (1)

Where:

\vec F_{1}, \vec F_{2}, \vec F_{3} - External forces exerted on the ring, measured in newtons.

\vec O - Vector zero, measured in newtons.

If we know that \vec F_{1} = (70.711,70.711)\,[N], \vec F_{2} = (-126.859, 46.173)\,[N], F_{3} = (F_{3,x},F_{3,y}) and \vec O = (0,0)\,[N], then we construct the following system of linear equations:

\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N (2)

\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N (3)

The solution of this system is:

F_{3,x} = 56.148\,N, F_{3,y} = -116.884\,N

The x-component of F_{3} is 56.148 newtons.

5 0
3 years ago
A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t square
torisob [31]

Answer:

a) 2.933 m

b) 4.534 m

Explanation:

We're given the equation

v(t) = -0.4t² + 2t

If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.

See attachment for the calculations

The conclusion of the attachment will be

7.467 - 2.933 and that is 4.534 m

Thus, The distance it travels in the second 2 sec is 4.534 m

6 0
3 years ago
A car with a mass of 850kg is moving at a speed of 72km/h when colliding with a concrete wall until it stops. After the collisio
Sergeeva-Olga [200]

Answer:

Explanation:

The vehicle is experiencing a large force created by the concrete wall.

Equation

vf^2 = vi^2 + 2*a * d

Givens

vf = 0   The car eventually does stop.

vi = 72 km/hr * [ 1000 m/  km] * [1 hour / 3600 seconds]

vi = 20 meters / second

a = ?

m = 850 kg

Solution

vf^2 = vi^2 + 2a*d

0 = 20 m/s + 2* 2 *a

-20 m/s = 4a

-20/4 =   a

a = - 5 m/s^2   The minus sign tells you the vehicle is slowing down. It sure should be.

Force = m * a

F = - 850 * (-5)

F =  - 4250 N

The car provides a 4250 N force on it going east to west and a 4250 N force going from west to east provided by the concrete wall.

8 0
3 years ago
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