Someone please help with these 2
jcherry99 avatar

Answers

Answer 1
Answer:

Answer:

Explanation:

The formula that you are working with is F = m*a

Since mass is one part of the formula if you increase the mass, you are going to increase the force.

The second one is much more difficult to answer because it is basically incomplete. This is one way to interpret it. If you start at a certain speed and increase during a known time period then effectively you are defining acceleration which is "a" in the formula.

Without those modifications, there is no answer.


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a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How far from the wall should the observer stand to get the best view?

Answers

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = (d)/(x)

\theta = tan^(-1) (d)/(x)

Now for the \DeltaBAC:

tan\theta = (d + h)/(x)

\theta = tan^(-1) (d + h)/(x)

Now, differentiating w.r.t x:

(d\theta )/(dx) = (d)/(dx)[tan^(-1) (d + h)/(x) -  tan^(-1) (d)/(x)]

For maximum angle, (d\theta )/(dx) = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = (-(d + h))/((d + h)^(2) + x^(2)) -(-d)/(x^(2) + d^(2))

(-(d + h))/((d + h)^(2) + x^(2)) = \frac{{d}{x^(2) + d^(2)}

After solving the above eqn, we get

x = \sqrt{(d)/(d + h)}

The observer should stand at a distance equal to x = \sqrt{(d)/(d + h)}

Final answer:

For optimum viewing of a painting in a gallery, an observer should position themselves a distance away from the painting calculated using Pythagoras theorem, forming a right-angled triangle with the painting and the floor. This distance can be expressed as c = √[(h/2 + d)² + (h/2)²], where h is the height of the painting and d is the height from the observer's eye to the bottom of the painting.

Explanation:

In the physics of optics, the viewer should position themselves to where they form a right-angled triangle with the ceiling and the painting leading to the best viewing experience. This is widely known as the 'normal viewing distance'.

Given that the painting has a height h and its lower edge is at a distance d above the observer's eye, the observer should stand a distance away from the wall, which can be calculated using Pythagoras' theorem in right triangles, which states that the square of the hypotenuse (c) is equal to the sum of the square of the other two sides (a and b), i.e., c² = a² + b²

Since the painting height and viewer height forms the right-angle in this case, we have: a = (h/2 + d), and b = h/2. Substituting a and b in Pythagoras equation, we can solve for c which is the required distance: c = √[(h/2 + d)² + (h/2)²]

Learn more about Optimum viewing distance here:

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Nina and Jon are practicing an ice skating routine. Nina is standing still. Jon, who is twice as heavy as Nina, skates toward her, pushing Nina away with force f. Assuming the system is closed, which statement is correct about this system? a. Nina experiences a force equal to f/2. b. Nina experiences a force equal to f^2. c. Nina experiences a force equal to 2f. d. Nina experiences a force equal to f.

Answers

Answer:

A

Explanation:

• Nina experiences a force equal to f.

Answer:

Nina experiences a force equal to f

Explanation:

got to get that 2nd answer slot correct too before an abusive expert verifier with an alt comes in and purposely verifies the wrong answer

Two point charges have a total electric potential energy of -24 J, and are separated by 29 cm.If the total charge of the two charges is 45 μC, what is the charge, in μC, on the positive one?
What is the charge, in μC, on the negative one?

Answers

Answer:

The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

Explanation:

Electric potential energy between two point charges is derived from concept of Work, Work-Energy Theorem and Coulomb's Law and described by the following formula:

U_(e) = (k\cdot q_(A)\cdot q_(B))/(r)(1)

Where:

U_(e) - Electric potential energy, measured in joules.

q_(A), q_(B) - Electric charges, measured in coulombs.

r - Distance between charges, measured in meters.

k - Coulomb's constant, measured in kilogram-cubic meters per square second-square coulomb.

If we know that U_(e) = -24\,J, q_(A) = 45* 10^(-6)\,C+ q_(B), k = 9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2)) and r = 0.29\,m, then the electric charge is:

-24\,J = -(\left(9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2)) \right)\cdot (45* 10^(-6)\,C+q_(B))\cdot q_(B))/(0.29\,m)

-6.96 = -405000\cdot q_(B)-9* 10^(9)\cdot q_(B)^(2)

9* 10^(9)\cdot q_(B)^(2)+405000\cdot q_(B) -6.96 = 0(2)

Roots of the polynomial are found by Quadratic Formula:

q_(B,1) = 1.327* 10^(-5)\,C, q_(B,2) \approx -5.827* 10^(-5)\,C

Only the first roots offer a solution that is physically reasonable. The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicle traveling 70 mi/hr?

Answers

Answer:

276.74* 10^8Mg/m^3

31.29 m/sec

Explanation:

We have given density of substance 0.14lb/in^3

We have convert this into Mg/m^3

We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg

We know that 1 kg = 1000 g ( 1000 gram )

So 0.06349 kg = 63.49 gram

And we know that 1 gram = 1000 milligram

So 63.49 gram =63.49* 10^3\ Mg

We know that 1 in^3=1.6387* 10^(-5)m^3

So 0.14in^3=0.14* 1.6387* 10^(-5)=0.2294* 10^(-5)m^3

So 0.14lb/in^3 =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]

In second part we have to convert 70 mi/hr to m/sec

We know that 1 mi = 1609.34 meter

So 70 mi = 70×1609.34 = 112653.8 meter

1 hour = 3600 sec

So 70 mi/hr =(70* 1609.34meter)/(3600sec)=31.29m/sec

A mouse is running across a room with a speed of 2.2 m/s. The mass of the mouse is 1.4 kg. What is the Kinetic Energy of the mouse?

Answers

Answer:

3.39 J

Explanation:

The kinetic energy of an object can be found by using the formula

k =  (1)/(2) m {v}^(2)  \n

m is the mass

v is the velocity

From the question we have

k =  (1)/(2)  * 1.4 *  {2.2}^(2)  \n  = 0.7 * 4.84 \n  = 3.388 \:  \:  \:  \:  \:  \:  \:  \:  \:

We have the final answer as

3.39 J

Hope this helps you

Answer:

im not sure

Explanation:

Hercules X-1 is a pulsating X-ray source. The X-rays from this source sometimes completely disappear for 6 hours every 1.7 days because the neutron star has a 1.7-day orbital period around its companion star, and it is eclipsed for ____ hours once every orbital period.

Answers

Answer:

06 Hours

Explanation:

As per the details given in the question it self, the neutron star X-1 is revolving around its companion star. The orbital period is 1.7 years which means it will complete the revolution in 1.7 years. During the movement in the orbit we will be able to detect the x-rays except for the time when it goes behind the companion star and eclipsed by it as seen from Earth.

Since the x-rays disappear completely for around 6 hours. This clearly means that eclipse period is 06 hours.

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