Ask question

Ask question

All categories

2 years ago

13

You are determining the mass of an object using a triple beam balance. When the pointer is lined up with the zero mark, the ride

rs show values of 300 grams, 20 grams, and 8.0 grams. What is the mass of the object?

2 answers:

astra-53 [7]2 years ago

3 0

We add the values on each of the riders:
300 + 20 + 8
= 328 grams

Setler [38]2 years ago

3 0

Answer:

167

Explanation:

You might be interested in

Balanced or Unbalanced?

Galina-37 [17]

Unbalanced because if it is pushing then stopping, that means that it is unbalanced.

6 0

2 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

6 0

2 years ago

A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark

Natali5045456 [20]

The total displacement of the toy car at the given positions is 0.

The given parameters;

<em>First displacement of the car, = 5 cm left</em>
<em>Second displacement of the car, = 8 cm right</em>
<em>Third displacement of the car, = 3 cm to the left</em>

The total displacement of the car is calculated as follows;

<em>Let the </em><em>left </em><em>direction be "</em><em>negative </em><em>direction"</em>
<em>Let the </em><em>right </em><em>direction be "</em><em>positive </em><em>direction"</em>

$\Delta x = - \ 5\ cm \ + \ 8 \ cm \ - \ 3 \ cm \\\\\Delta x = 0$

Thus, the total displacement of the toy car at the given positions is 0.

Learn more about displacement here: brainly.com/question/18158577

6 0

1 year ago

Read 2 more answers

What's ur fav song?<br><br> Fr33 Pointtsss <3

arsen [322]

Answer:

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Haha Rick rolled you

Explanation:

jk my favourite song is Thunder, Despacito

3 0

2 years ago

Read 2 more answers

Two identical charges, 2 m apart, exert forces of magnitude 4 N on each other. The value of each charge is: 1. 9 × 105 C 2. 4.2

lesya692 [45]

Answer:

The value of each charge is 4.22 x 10⁻⁵ C

Explanation:

Given;

distance between the two identical charges, d = 2 m

the force of repulsion between these two charges, F = 4N

Apply Coulomb's law;

$F = \frac{kq_1q_2}{r^2} \\\\but \ q_1 =q_2,then \ let \ q_1 =q_2 = q\\\\F = \frac{kq^2}{r^2}\\\\q^2 = \frac{Fr^2}{k}\\\\q^2 = \frac{4*2^2}{9*10^9} \\\\q ^2 = 1.7778*10^{-9}\\\\q = \sqrt{1.7778*10^{-9}}\\\\q =4.22 *10^{-5} C\\\\q= q_1=q_2= 4.22 *10^{-5} C$

Therefore, the value of each charge is 4.22 x 10⁻⁵ C

7 0

2 years ago