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2 years ago

Explain how rock D is the best rock

Physics

1 answer:

Naya [18.7K]2 years ago

7 0

The best rock is D because it is big and heavy out of all of them. Rock A would fling too far because it is the lightest of all rocks as shown.

B and C would probably fling through the target because it is just the right size.

Besides, Depending on their weight of each rock, rock D is heavy because it is big. Rock A is too light to fling or land on the target. This is my hypothesis.

I don't really get how would rock D be the best one, but upon your question, this is all i know.

Thank you for posting your question, hope this helps, have a great day. please mark brainliest so i can answer more of your questions.

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What is the law of conservation of energy?

charle [14.2K]

The law of conservation of energy, a fundamental concept of physics, states that the total amount of energy remains constant in an isolated system. It implies that energy can neither be created nor destroyed, but can be change from one form to another.

6 0

2 years ago

Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.

aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

6 = (20 / 32.2 + 50 / 32.2)a

2.173a = 6

A = 2.76 ft/s^2

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2

6 0

2 years ago

Describe how elements are formed in stars.

elena-14-01-66 [18.8K]

Answer:

Stars create new elements in their cores by squeezing elements together in a process called nuclear fusion.

Explanation:

6 0

2 years ago

A laser beam is incident at an angle of 33.0° to the vertical onto a solution of cornsyrup in water.(a) If the beam is refracted

IrinaK [193]

Answer:

1.29649

488.08706 nm

$6.14644\times 10^{14}\ Hz$

231715700.28346 m/s

Explanation:

n denotes refractive index

1 denotes air

2 denotes solution

$\lambda_0$ = 632.8 nm

From Snell's law we have the relation

$n_1sin\theta_1=n_2sin\theta_2\\\Rightarrow n_2=\dfrac{n_1sin\theta_1}{sin\theta_2}\\\Rightarrow n_2=\dfrac{1\times sin33}{sin24.84}\\\Rightarrow n_2=1.29649$

Refractive index of the solution is 1.29649

Wavelength is given by

$\lambda=\dfrac{\lambda_0}{n_2}\\\Rightarrow \lambda=\dfrac{632.8}{1.29649}\\\Rightarrow \lambda=488.08706\ nm$

The wavelength of the solution is 488.08706 nm

Frequency is given by

$f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{488.08706\times 10^{-9}}\\\Rightarrow f=6.14644\times 10^{14}\ Hz$

The frequency is $6.14644\times 10^{14}\ Hz$

$v=\dfrac{c}{n_2}\\\Rightarrow v=\dfrac{3\times 10^8}{1.29469}\\\Rightarrow v=231715700.28346\ m/s$

The speed in the solution is 231715700.28346 m/s

8 0

2 years ago

Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.

Ksivusya [100]

$|v| =\sqrt{ G \cdot M / r}$ , where

$M$ the mass of the planet, and
$G$ the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

Equation 1 (see below) relates net force the object experiences, $\Sigma F$ to its orbit velocity $v$ and its mass $m$ required for it to stay in orbit :

$\Sigma F = m \cdot v^{2} / r$ (equation 1)

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force $\Sigma F$ acting on the soccer ball shall equal to its weight, $W = m \cdot g$ where $g$ the gravitational acceleration constant. Thus

$\Sigma F = W = m \cdot g$ (equation 2)

Substitute equation 2 to the left hand side of equation 1 and solve for $v$ ; note how the mass of the soccer ball, $m$ , cancels out:

$m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0)$ (equation 3)

Equation 4 gives the value of gravitational acceleration, $g$ , a point of negligible mass experiences at a distance $r$ from a planet of mass $M$ (assuming no other stellar object were present)

$g = G \cdot M/ r^{2}$ (equation 4)

where the universal gravitation constant $G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}$

Thus

$\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}$

3 0

3 years ago