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Kaylis [27]
3 years ago
7

When an unbalanced force acts on an object

Physics
1 answer:
xz_007 [3.2K]3 years ago
3 0

Answer:

D. The object accelerates

Explanation:

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The following acceleration vs time plots show
fenix001 [56]

The data set that is most likely from the car equipped with ABS is :

  • The First ( upper ) plot

<u><em>Although the plots related to your question is missing attached below is the missing data.  </em></u>

The First plot ( acceleration vs time ) plot is the data that represents the application of brakes in a car fitted with ABS ( antilock braking system ) because of the absence of skidding effect between 6.5 and 9 seconds in the graph.

The skidding effect is present in the second plot around the same time because the car is not fitted with ABS  

Hence we can conclude that the graph that shows the car fitted with ABS is the first plot .

Learn more : brainly.com/question/4360615

7 0
3 years ago
Why is gravitational forces equal for everyone?
Alik [6]

That equation is Newton's universal law of gravitation. ... Any two masses exert equal-and-opposite gravitational forces on each other. If we drop a ball, the Earth exerts a gravitational force on the ball, but the ball exerts a gravitational force of the same magnitude (and in the opposite direction) on the Earth.

7 0
4 years ago
Use the information below to answer questions
Ulleksa [173]

Answer:

The charges are q₁  = 2 × 10⁻⁸ C and  q₂ = 3 × 10⁻⁸ C

Explanation:

Here is the complete question

Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.

Solution

The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by

F = kq₁q₂/r₁²

q₁q₂ = Fr₁²/k

q₁q₂ = 1.35 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.054/9 × 10⁻¹³ C² = 0.006 × 10⁻¹³ C² = 6 × 10⁻¹⁶ C²

q₁q₂ = 6 × 10⁻¹⁶ C² (1)

When the charges are brought together, the charge is now q = (q₁ + q₂)/2

The new repulsive force F = 1.406 × 10⁻⁴ N  at a distance of r₂ = 20 cm = 0.2 m is then

F₂ = kq²/r₂²

q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²

q² = 6.25 × 10⁻¹⁶ C²

q = √(6.25 × 10⁻¹⁶) C

q = 2.5 × 10⁻⁸ C

(q₁ + q₂)/2 =  2.5 × 10⁻⁸ C

(q₁ + q₂) = 2 × 2.5 × 10⁻⁸ C

q₁ + q₂ = 5 × 10⁻⁸ C (2)

q₁  = 5 × 10⁻⁸ C - q₂  (3)

Substituting equation (3) into (1), we have

(5 × 10⁻⁸ C - q₂)q₂ = 6 × 10⁻¹⁶ C²

Expanding the bracket, we have

(5 × 10⁻⁸ C)q₂ - q₂² = 6 × 10⁻¹⁶ C²

So, q₂² - (5 × 10⁻⁸ C)q₂ + 6 × 10⁻¹⁶ C² = 0

Using the quadratic formula to find q₂

q_{2} = \frac{-(-5 X 10^{-8} )+/- \sqrt{(-5 X 10^{-8} )^{2} - 4X1X6 X 10^{-16} } }{2X1}\\  = \frac{5 X 10^{-8} )+/- \sqrt{25 X 10^{-16}  - 24 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- \sqrt{1 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- 1 X 10^{-8} }{2}\\= \frac{5 X 10^{-8} + 1 X 10^{-8} }{2} or \frac{5 X 10^{-8}  - 1 X 10^{-8} }{2}\\= \frac{6 X 10^{-8} }{2} or \frac{4 X 10^{-8}}{2}\\= 3 X 10^{-8} C or 2 X 10^{-8} C

q₁  = 5 × 10⁻⁸ C - q₂

q₁  = 5 × 10⁻⁸ C - 3 × 10⁻⁸ C or 5 × 10⁻⁸ C - 2 × 10⁻⁸ C

q₁  = 2 × 10⁻⁸ C or 3 × 10⁻⁸ C

So the charges are q₁  = 2 × 10⁻⁸ C and  q₂ = 3 × 10⁻⁸ C

5 0
4 years ago
Familiarize yourself with the map showing the DSDP Leg 3 drilling locations and the position of the mid-ocean ridge (Figure 1 to
Inga [223]

Answer:

For more than 40 years, results from scientific ocean drilling have contributed to global understanding of Earth’s biological, chemical, geological, and physical processes and feedback mechanisms. The majority of these internationally recognized results have been derived from scientific ocean drilling conducted through three programs—the Deep Sea Drilling Project (DSDP; 1968-1983), the Ocean Drilling Program (ODP; 1984-2003), and the Integrated Ocean Drilling Program (IODP; 2003-2013)—that can be traced back to the first scientific ocean drilling venture, Project Mohole, in 1961. Figure 1.1 illustrates the distribution of drilling and sampling sites for each of the programs, and Appendix A presents tables of DSDP, ODP, and IODP legs and expeditions. Although each program has benefited from broad, international partnerships and research support, the United States has taken a leading role in providing financial continuity and administrative coordination over the decades that these programs have existed. Currently, the United States and Japan are the lead international partners of IODP, while a consortium of 16 European countries and Canada participates in IODP under the auspices of the European Consortium for Ocean Research Drilling (ECORD). Other countries (including China, Korea, Australia, New Zealand, and India) are also involved.

As IODP draws to a close in 2013, a new process for defining the scope of the next phase of scientific ocean drilling has begun. Illuminating Earth’s Past, Present, and Future: The International Ocean Discovery Program Science Plan for 2013-20231 (hereafter referred to as “the science plan”), which is focused on defining the scientific research goals of the next 10-year phase of scientific ocean drilling, was completed in June 2011 (IODP-MI, 2011). The science plan was based on a large, multidisciplinary international drilling community meeting held in September 2009.2 A draft of the plan was released in June 2010 to allow for additional comments from the broader geoscience community prior to its finalization. As part of the planning process for future scientific ocean drilling, the National Science Foundation (NSF) requested that the National Research Council (NRC) appoint an ad hoc committee (Appendix B) to review the scientific accomplishments of U.S.-supported scientific ocean drilling (DSDP, ODP, and IODP) and assess the science plan’s potential for stimulating future transformative scientific discoveries (see Box 1.1 for Statement of Task). According to NSF, “Transformative research involves ideas, discoveries, or tools that radically change our understanding of an important existing scientific or engineering concept or educational practice or leads to the creation of a new paradigm or field of science, engineering, or education. Such research challenges current understanding or provides pathways to new frontiers.”3 This report is the product of the committee deliberations on that review and assessment.

HISTORY OF U.S.-SUPPORTED SCIENTIFIC OCEAN DRILLING, 1968-2011

The first scientific ocean drilling, Project Mohole, was conceived by U.S. scientists in 1957. It culminated in drilling 183 m beneath the seafloor using the CUSS 1 drillship in 1961. During DSDP, Scripps Institution of Oceanography was responsible for drilling operations with the drillship Glomar Challenger. The Joint Oceanographic Institutions for Deep Earth Sampling (JOIDES), which initially consisted of four U.S. universities and research institutions, provided scientific advice. Among its numerous achievements, DSDP

Explanation:

7 0
3 years ago
Help please! Will give brainliest to first correct answer!
Olegator [25]

Answer:

1) Addition of a catalyst

2) To change the reaction rate of slope B to look like slope A, simply add a catalyst to speed up the rate of reaction, giving you a higher amount of products in a shorter amount of time (line A)

Explanation:

1 and 2)Two things can alter the rate of a reaction, either the addition of a catylist which will not alter the composition of the products or reactants, but will accelerate the reaction time, or an increase in temperature will also increase the rate at which a reaction will occur.

You could choose temperature also and have the same result, it's your choice both are correct, but catalyst is the easiest.

   

8 0
3 years ago
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