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3 years ago

All digits shown on measuring device, plus one estimated digit, are considered_______.

Physics

1 answer:

vichka [17]3 years ago

3 0

Significant

Explanation:

All the digits shown on a measuring device and one estimated digit are all considered to be significant.

A significant digit is a set of values that shows how precise a measurement is reported.

Measuring devices such as calculators gives their values in significant digits.

Non- zero digits in a measurement are always significant.
Zero's before a decimal are not significant. Those after a number in a decimal are significant.
Zero's between digits are significant.

learn more:

Significant digits brainly.com/question/2743055

#learnwithBrainly

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Options: A.) 10 N B.) 15 N C.) 25 N D.) 35N

Lady_Fox [76]

Given:

F_gravity = 10 N

F_tension = 25 N

Let's find the net centripetal force exterted on the ball.

Apply the formula:

$\sum ^{}_{}F_{\text{net}}=F_1+F_2=F_{centripetal}$

From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.

Hence, the tensional force is positive while the gravitational force is negative.

Thus, we have:

$F_{\text{net}}=F_{\text{centripetal}}=F_{tension}-F_{gravity}=25N-10N=15N$

Therefore, the net centripetal force exterted on the ball is 15 N.

ANSWER:

15 N

7 0

1 year ago

1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (H

JulijaS [17]

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.

The electric field lines are always perpendicular to the equipotential surface.

$dV = \overrightarrow{E} . d\overrightarrow{r}\\\\$

For equipotential surface, dV = 0 so

$0 = \overrightarrow{E} . d\overrightarrow{r}\\\\$

The dot product of two non zero vectors is zero, if they are perpendicular to each other.

5 0

3 years ago

On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const

IceJOKER [234]

Answer:

Explanation:

Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} (1)$

Since the car starts from rest, v₀ =0.
We know the value of t = 5 sec., but we need to find the value of a.
Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

$a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2 (2)$

Replacing a and t in (1):

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m. (3)$

Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

$a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2 (4)$