1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
tankabanditka [31]
2 years ago
13

Two objects gravitationally attract with a force of 100 N. If the mass of one object is doubled and the mass of the other object

.is tripled,
what is the new gravitational force?
Physics
1 answer:
barxatty [35]2 years ago
5 0

Hello!

Recall the equation for gravitational force:
F_g = \frac{Gm_1m_2}{r^2}

Fg = Force of gravity (N)
G = Gravitational constant

m1, m2 = masses of objects (kg)
r = distance between the objects' center of masses (m)

There is a DIRECT relationship between mass and gravitational force.

We are given:
F_g = 100N

If we were to double one mass and triple another, according to the equation:
F'_g = \frac{G(2m_1)(3m_2)}{r^2} = 6(\frac{G(m_1)(m_2)}{r^2}) = 6F_g

Thus:
6 * F_g = 6 * 100 = \boxed{600N}

You might be interested in
Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a s
PSYCHO15rus [73]

Answer:

Explanation:

wave length of light λ = 502 nm

screen distance D = 1.2 m

width of one fringe = 10.2 mm / 20

= .51 mm

fringe width = λ D / a  , a is separation of slits

Puting the values given

.51 x 10⁻³ =  502 x 10⁻⁹ x 1.2 / a

a = 502 x 10⁻⁹ x 1.2 / .51 x 10⁻³

= 1181.17 x 10⁻⁶ m

1.18 x 10⁻³ m

= 1.18 mm .

8 0
2 years ago
Suppose a ray of light traveling in a material with an index of refraction n a reaches an interface with a material having an in
KATRIN_1 [288]

Answer: C and D

Explanation: One of the first rule for total internal reflection to occur is that the ray must move from a dense to a less dense medium, hence refractive index of medium a must be greater than that of b.

When a ray moves from a dense to a less dense medium, the refracted ray moves away from the normal thus increasing the size of the angle of refraction (total internal refraction occurs when the angle of refraction is 90° and the angle of incidence at this point is known as the critical angle), hence the angle of incidence must be greater than the critical angle.

These points verifies option C and D

5 0
3 years ago
A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

7 0
3 years ago
A black widow spider hangs motionless from a web that extends vertically from the ceiling above. If the spider has a mass of 1.2
leva [86]

Answer:

Tension = 0.012 N

Explanation:

If the black widow spider is hanging vertically motionless from the ceiling above. Then, the weight of the spider must be balancing the tension in the spider web. Therefore,

Tension = Weight

Tension = mg

where,

m = mass of spider = 1.27 g = 0.00127 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

Tension = (0.00127 kg)(9.8 m/s²)

<u>Tension = 0.012 N</u>

8 0
3 years ago
6. A car moving in a straight line at constant speed:
ddd [48]

Answer:

D. has no overall force acting on it.

Explanation:

Why?

Because in a straight line at the constant speed means the car moving in the same velocity, which is not acceleration neither deceleration, and it cannot be on a downhill slope. So the correct answer is

<h3>→ D. has no overall force acting on it.</h3>
8 0
3 years ago
Other questions:
  • Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. spec
    15·1 answer
  • Which factors are most likely to limit the growth of the coral reef in the caribbean?
    5·2 answers
  • How would a bar magnet orient itself relative to this planet's magnetic field
    9·2 answers
  • The most common purpose for fracking in the u.s is to get
    12·1 answer
  • A chemist has two compounds that do not chemically react. He thinks that by finding the right catalyst, he can get them to react
    15·1 answer
  • Name four factors that affect local, regional climates?
    5·2 answers
  • 14. Heat energy equal to 25,000 J is applied to a 1200 g brick whose specific heat is 2.45 J/gºC.
    14·1 answer
  • Which term describes the number of thoughs that pass a point in a given amount of time
    10·1 answer
  • Two identical circular, wire loops 35.0 cm in diameter each carry a current of 2.80 A in the same direction. These loops are par
    13·1 answer
  • What force is affected when the distance between two objects remains the same and the mass of each object is doubled?.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!