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3 years ago

A) Find the displacement for the given time intervals ,

Physics

2 answers:

klasskru [66]3 years ago

6 0

Answer:

the answer is that it will be positive

Olegator [25]3 years ago

3 0

Answer:

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Explanation:

vsgshzjdjdk

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The precision of a laboratory instrument is ± 0.05 g. The accepted value for your measurement is 7.92 g. Which measurements are

skelet666 [1.2K]

Answer:

7.89 7.91

Explanation:

The ranges of measurement lie between 7.92-0.05 and 7.92+0.05

7.87g and 7.97g

3 0

3 years ago

Read 2 more answers

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp

laila [671]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

$a = \frac{v^2}{r}$

At constant speed, we will have;

$v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1$

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0

3 years ago

Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als

lesantik [10]

Answer:

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

(As the positive charge would move away from positive charges and would be attracted by negative ones).

So, the combination of answers that is true is b) (positive, negative, positive).

3 0

3 years ago

What is the internal energy of 2.00 mol of diatomic hydrogen gas (H2) at 35°C?

djyliett [7]

As you mentioned, we will use Equipartition Theorem.
H2 has 5 degrees of freedom; 3 translations and 2 rotation
Therefore:
Internal energy = (5/2) nRT
You just substitute in the equation with the values of R and T and calculate the internal energy as follows:
Internal energy = (5/2) x 2 x 8.314 x 308 = 32.0089 x 10^3 J

4 0

3 years ago

A race car has a centripetal acceleration of 15.625 m/s2 as it goes around a curve. If the curve is a circle with radius 40 m, w

myrzilka [38]

The centripetal acceleration is given by
$a_c = \frac{v^2}{r}$
where v is the tangential speed and r the radius of the circular orbit.

For the car in this problem, $a_c = 15.625 m/s^2$ and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
$v= \sqrt{a_c r}= \sqrt{(15.625 m/s^2)(40 m)}=25 m/s$

8 0

3 years ago