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Aleonysh [2.5K]
3 years ago
12

SOS HELP ME

Physics
1 answer:
Dafna1 [17]3 years ago
5 0

Answer:

Angle Vy Vy^2/(2*g)

25         6.3    2.02

30          7.5  2.87

45         10.6 5.73

50         11.5 6.75

65         13.6 9.44

70         14.1 10.1

Vy = vertical velocity = V sin theta

2 g H = Vy^2      since the final velocity is zero

H = Vy^2 / (2 * g)

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A 10n force is applied to a 25kg mass to slide it across a frictional surface. What is the acceleration of the mass?
klasskru [66]

Answer: a = 0.4m/s^2 - 9.8*c where c is the coefficient of kinetic friction of the surface

Explanation: We know that, by the second Newton's law, a = F/m

where a is the acceleration, F is the net force and m is the mass of the object.

Then, if the surface is frictionless, the total force applied in the object is 10N, and the mass of the object is 25kg, so the acceleration is:

a =10N/25kg = 0.4m/s^2.

But if the surface is frictional, there will be a force of friction applied in the mass (this depends on the coefficient of friction and the weight of the mass), this means that the acceleration will be reduced.

If = -(9.8*25)*c

where c is a number that is bigger than 0 and smaller than 1, is called the coefficient of kinetic friction.

So the total force is now:

F = (10 - 9.8*25*c)

Then, the acceleration in a frictional surface is equal to:

a = (10 - 9.8*25*c)/25 = 0.4m/s^2 - 9.8*c

6 0
3 years ago
Read 2 more answers
A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i
Arturiano [62]

Answer:

q / q_{1} = 2, assuming that q_{1} and (q - q_{1}) are point charges.

Explanation:

Let k denote the coulomb constant. Let r denote the distance between the two point charges. In this question, neither k and r depend on the value of q_{1}.

By Coulomb's Law, the magnitude of electrostatic force between q_{1} and (q - q_{1}) would be:

\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}.

Find the first and second derivative of F with respect to q_{1}. (Note that 0 < q_{1} < q.)

First derivative:

\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}.

Second derivative:

\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}.

The value of the coulomb constant k is greater than 0. Thus, the value of the second derivative of F with respect to q_{1} would be negative for all real r. F\! would be convex over all q_{1}.

By the convexity of \! F with respect to \! q_{1} \!, there would be a unique q_{1} that globally maximizes F. The first derivative of F\! with respect to q_{1}\! should be 0 for that particular \! q_{1}. In other words:

\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0<em>.</em>

2\, q_{1} = q.

q_{1} = q / 2.

In other words, the force between the two point charges would be maximized when the charge is evenly split:

\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}.

3 0
2 years ago
How would you describe the behavior of particles in a solid?
Minchanka [31]
How would you describe the behavior of particles in a solid?
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3 years ago
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What will make two bar magnets repel each<br> other?
Kitty [74]
The arrows always start at the magnet's north pole and point towards its south pole. When two like-poles point together, the arrows from the two magnets point in OPPOSITE directions and the field lines cannot join up. So the magnets will push apart
6 0
2 years ago
A fan that is rotating at 960 rev/s is turned off. It makes 1500 revolutions before it comes to a stop. a) What was its angular
Evgesh-ka [11]

Answer:

α = 1930.2 rad/s²

Explanation:

The angular acceleration can be found by using the third equation of motion:

2\alpha \theta=\omega_f^2-\omega_i^2

where,

α = angular acceleration = ?

θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s

Therefore,

2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}

<u>α = - 1930.2 rad/s²</u>

<u>negative sign shows deceleration</u>

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