2.

During an off-day, Mike Trout, Shohei Ohtani, and Albert Pujols and having a pool party. They are tugging on a ring for some odd

andrew11 [14]

Answer:

$\left \| F \right \| = 42.3 N$

$\theta = 11.7$ degree north of west

Explanation:

Given data:

Force by trout 150 N

Force by Ohtani 100 N

Force by pujols 200 N

Net force is given as

$F = 150 j - 200\times cos 45j + ( 100i - 200\times sin45 i)$

$F = (100 - 100\sqrt {2}) i + (150 - 100\sqrt{2}) j$

$\left \| F \right \| = \sqrt{(100 - 100\sqrt{2})^2 + (150 -100\sqrt{2})^2}$

$\left \| F \right \| = 42.3 N$

Direction is

$tan\theta = \left | \frac{150-100\sqrt{2}}{100-100\sqrt{2}} \right |$

$\theta = 11.7$ degree north of west

6 0

4 years ago

A block of mass M is connected by a string and pulley to a hanging mass m.

sveticcg [70]

(b) Use Newton's second law. The net forces on block M are

• ∑ F (horizontal) = T - f = Ma … … … [1]

• ∑ F (vertical) = n - Mg = 0 … … … [2]

where T is the magnitude of the tension, f is the mag. of kinetic friction between block M and the table, a is the acceleration of block M (but since both blocks are moving together, the smaller block m also shares this acceleration), and n is the mag. of the normal force between the block and the table.

Right away, we see n = Mg, and so f = µn = 0.2Mg.

The net force on block m is

• ∑ F = mg - T = ma … … … [3]

You can eliminate T and solve for a by adding [1] to [3] :

(T - 0.2Mg) + (mg - T ) = Ma + ma

(m - 0.2M) g = (M + m) a

a = (10 kg - 0.2 (20 kg)) (9.8 m/s²) / (10 kg + 20 kg)

a = 1.96 m/s²

We can get the tension from [3] :

T = m (g - a)

T = (10 kg) (9.8 m/s² - 1.96 m/s²)

T = 78.4 N

(c/d) No time duration seems to be specified, so I'll just assume some time t before block M reaches the edge of the table (whatever that time might be), after which either block would move the same distance of

1/2 (1.96 m/s²) t

(e) Assuming block M starts from rest, its velocity at time t is

(1.96 m/s²) t

(f) After t = 1 s, block M reaches a speed of 1.96 m/s. When the string is cut, the tension force vanishes and the block slows down due to friction. By Newton's second law, we have

∑ F = -f = Ma

The effect of friction is constant, so that f = 0.2Mg as before, and

-0.2Mg = Ma

a = -0.2g

a = -1.96 m/s²

Then block M slides a distance x such that

0² - (1.96 m/s²) = 2 (-1.96 m/s²) x

x = (1.96 m/s²) / (2 (1.96 m/s²))

x = 0.5 m

(I don't quite understand what is being asked by the part that says "calculate the time taken to contact block M and pulley" …)

Meanwhile, block m would be in free fall, so after 1 s it would fall a distance

x = 1/2 (-9.8 m/s²) (1 s)

x = 4.9 m

7 0

3 years ago

Which mass is the same as 0.031 g?

LenaWriter [7]

0.031 g is equal to 31 grams

6 0

3 years ago

Read 2 more answers

Which statement is the best interpretation of the ray diagram shown?

n200080 [17]

Answer:

Explanation:

First off, this lens is concave. Second, the image is obviously smaller, and third, the only thing that is NOT obvious, is the fact that real images are always upside down and virtual images are always right-side-up. So the choice you're looking for is D.

6 0

3 years ago

How are atomic emission spectra like fingerprints for the elements

storchak [24]

Atomic emission spectra are like fingerprints for the elements, because it can show the number of orbits in that elements as well as the energy levels of that element. As each emission of atomic spectra is unique, it is the fingerprint of element.

Explanation:

Each element has unique arrangement of electrons in different energy levels or orbits. So depending upon the difference in energy of the orbital, the emission spectra will be varying for each element. As the binding energy and excitation energy is not common for any two elements, so the spectra obtained when those excited electrons will release energy to ground state will also be unique.

As in atomic emission spectra, the incident light will be absorbed by the electrons of those elements making the electron to excite, then the excited electron will return to ground state on emission of radiation of energy. Thus, this energy of emission is equal to the difference between the energy of initial and final orbital. So the spectra will act like fingerprints for elements.

8 0

3 years ago