Ask question

Ask question

All categories

marishachu [46]

3 years ago

11

Multiply the following three numbers and report your answer to the correct number of significant figures: 0.020cm x 50cm x 11.1c

m=?

1 answer:

Neko [114]3 years ago

5 0

Answer:11.1

Explanation:

Three significant figures

You might be interested in

collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitu

anzhelika [568]

Answer:

metre per seconds

Explanation:

because velocity = distance ÷ time

4 0

3 years ago

When a 12 N horizontal force is applied to a box on a horizontal tabletop, the box remains at rest. The force of friction acting

Mrrafil [7]

Answer:

12N

Explanation:

when a force is applied to a body but still stays at rest or moves at a constant speed , the frictional force is equal to the force applied

3 0

2 years ago

Which of the following would produce the most power?

Fantom [35]

Answer:

A mass of 10 kilograms lifted 10 meters in 5 seconds.

Explanation:

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

$Power = \frac {Energy}{time}$

But Energy = mgh

Substituting into the equation, we have

$Power = \frac {mgh}{time}$

Given the following data;

Mass = 10kg

Height = 10m

Time = 5 seconds

We know that acceleration due to gravity is equal to 9.8 m/s²

$Power = \frac {10*9.8*10}{5} = 490 Watts$

Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.

6 0

2 years ago

The standard wave format for any wave is

Novay_Z [31]

The standard wave format for any wave is transverse wave

8 0

3 years ago

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago