Ask question

Ask question

All categories

3 years ago

7

A box experiencing a gravitational force of 600 N is being pulled to the right with force of 250 N. A 25 N frictional force acts

on the box as it moves to the right.

1 answer:

bearhunter [10]3 years ago

6 0

Answer: 0 NEWTONS

Explanation:

You might be interested in

The left end of a rod of length and rotational inertia is attached to a frictionless horizontal surface by a frictionless pivot,

natima [27]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

6 0

3 years ago

Use the mass spectrum of europium to determine the atomic mass of europium. where the peak representing eu-151 has an exact mass

slavikrds [6]

Answer: The atomic mass of a Europium atom is 151.96445 amu.

From the given information:

Percent intensity is 91.61% of Europium atom of molecular weight 150.91986 amu.

Percent intensity is 100.00% of Europium atom of molecular weight 152.92138 amu.

Abundance of Eu-151 atom:

$X_{Eu-151}=\frac{0.9161}{0.9161+1.000}=0.4781$

Abundance of Eu-153 atom:

$X_{Eu-153}=\frac{1.000}{0.9161+1.000}=0.5219$

Atomic mass of Europium atom:

$A=(X_{Eu-151}\times150.91986+X_{Eu-153}\times152.92138)amu\\A=(0.4781\times150.91986+0.5219\times152.92138)amu=151.96445 amu$

Therefore, the atomic mass of a Europium atom is 151.96445 amu.

3 0

3 years ago

Read 2 more answers

1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s

siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

4 0

3 years ago

A car moves 20 km towards the North and then 35 km at an angle of 60o towards west of North. Its magnitude of displacement from

Law Incorporation [45]

Answer:

15

Explanation:

displacement = initial position - final position

8 0

2 years ago

You drop a rock down a well and hear a splash 3 s later. As Charlie Brown would say, the well is 3 seconds deep. How fast is the

vampirchik [111]

Since it was dropped, it should be the speed of gravity which is 9.8 meters/second

5 0

2 years ago