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Liula [17]
3 years ago
7

A bullet that has been too badly mutilated to mark can be

Physics
1 answer:
Pavel [41]3 years ago
3 0
What are the options

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the object on the right gained mass so that it had as much mass as the object on the left, how would the gravitational force bet
kherson [118]

The gravitational force between the objects A. It would increase.

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m_1,m_2 are the masses of the two objects

r is the separation between the objects

In this problem, we are told that one of the object (the one on the right) gains mass: this means that, for instance, the value of m_2 increases. We can see from the equation that the gravitational force is directly proportional to the masses: therefore, if one of the masses increases (while the distance between the two objects remains constant), it means that the force also increases.

Therefore, the correct answer is

A. It would increase.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0
3 years ago
Read 2 more answers
Scientists communicate results in a variety of ways that include but are not limited to published scientific journals and period
RUDIKE [14]
S~ cientists publish their original research in scientific journals, which are fundamentally different from news magazines. The articles in scientific journals are not written by journalists – they are written by scientists. Scientific articles are not sensational stories intended to entertain the reader with an amazing discovery, nor are they news stories intended to summarize recent scientific events, nor even records of every successful and unsuccessful research venture. Instead, scientists write articles to describe their findings to the community in a transparent manner.
5 0
3 years ago
A 5 cm spring is suspended with a mass of 3.8589 g attached to it which extends the spring by 1.5747 cm. The same spring is plac
lana66690 [7]

Answer:

charges of the beads is 1.173 ×10^{-15} C

Explanation:

given data

mass = 3.8589 g = 0.003859 kg

spring length = 5 cm = 0.05 m

extend spring x = 1.5747 cm = 0.15747 m

spring's extension = 0.0116 m

to find out

charges of the beads

solution

we know that force is

force = mass × g

force = 0.003859 × 9.8

force = 0.03782 N

so we know  force for mass

force  = -kx

so k = force / x

put here force and x value

k = -0.03782 / 0.1575

k = -0.24 N/m

and

force for spring's extension

force = -kx

force = -0.24 ( 0.0116) = 0.002784 N

so here

total length L = 0.05 + 0.0116 = 0.0616

so charges of the beads = force × L² / ke

charges of the beads = 0.002784 × (0.0616)² / (9 ×10^{9} )

so charges of the beads = 1.173 ×10^{-15} C

3 0
3 years ago
If the result of your calculation of a quantity has si units kg•m^2/s^2•C, that quantity could be
Novosadov [1.4K]
It would be Joules.
Workdone is measured in Joules.
Workdone = Force * distance
Force = mass * acceleration
          = kg      *  ms⁻²
          = kgms⁻²

Distance = m

So, Force * distance
       kgms⁻² * m

Apply laws of indices that says
x² * x³ = x²⁺³ = x⁵

Therefore, It would be kgm²s⁻²
m¹ * m¹ = m¹⁺¹ = m²
s⁻² is also = s / 2





4 0
3 years ago
Read 2 more answers
A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
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