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sineoko [7]
3 years ago
7

EMERGENCY!! Doing my Physics homework now and I need help on this one please - all information provided, I'd be very VERY gratef

ul for help!! 15 points!!

Physics
1 answer:
IrinaK [193]3 years ago
6 0

(a) The vertical motion is accelerated by gravity. The horizontal component is constant (neglecting air resistance, if this is not a projectile motion, the horizontal component would also be accelerated)

(b) Vertical:

v_y = 30\sin 40^\circ = 19.3\frac{m}{s}

Horizontal:

v_x = 30\frac{m}{s}\cos 40^\circ = 23.0 \frac{m}{s}

(c) Use the kinematic equation for distance. Calculate only the vertical component (horizontal is irrelevant):

s_y = -\frac{1}{2}gt^2 +vt\,,\,\,\,s_y=0\\0 = -\frac{1}{2}gt^2 +vt\\\frac{1}{2}gt =v\\t = \frac{2v}{g}= \frac{2\cdot30\sin 40^\circ\frac{m}{s}}{9.8\frac{m}{s^2}}=3.9s

The ball will be in the air for about 3.9 s.

(d) The range is the horizontal distance traveled. We know the ball is in the air for 3.9s and it moves with a horizontal velocity of 23 m/s. So:

s_x = 23\frac{m}{s}\cdot 3.9 s = 89.7m

The range is 89.7 meters.

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A 10.0-cm-long uniformly charged plastic rod is sealed inside a plastic bag. The net electric flux through the bag is 7.50 × 10
Rina8888 [55]

Answer:

66.375 x 10⁻⁶ C/m

Explanation:

Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as ;

∅ = Q / ε₀        -----------------(i)

Where;

∅ = 7.5 x 10⁵ Nm²/C

ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²

Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;

7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)

Solve for Q;

Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²

Q = 66.375 x 10⁻⁷ C

Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e

L = Q / l     ----------------------(ii)

Where;

Q = 66.375 x 10⁻⁷ C

l = length of the rod = 10.0cm = 0.1m

Substitute these values into equation (ii) as follows;

L = 66.375 x 10⁻⁷C / 0.1m

L = 66.375 x 10⁻⁶ C/m

Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.

3 0
3 years ago
You whirl a stone in a horizontal circle in such a way that the stone is in uniform circular motion. Which of the following is t
horsena [70]

a. The direction of the stone's velocity changes as it moves around the circle.

b. The magnitude of the stone's velocity does not change.

d. The change in direction of the stone's motion is due to the centripetal force acting on the stone.

Above given are true for the given situation.

<u>Answer:</u> Option A, B and D

<u>Explanation:</u>

Circular motion may be characterized as the moving of an objects along the diameter of the circle or any circular direction. It may be standardized and non-uniform based on whether or not the rate of rotation is unchanged.

The velocity, a vector quantity is constant in a uniform circle motion speed is constant as its direction continues to change. Centripetal force works inward toward the core to counterbalance the centrifugal force from the center moving outward.

4 0
3 years ago
Arrange the stars based on their temperature. Begin with the coolest star, and end with the hottest star.
Slav-nsk [51]

Answer:

Uranus, Pluto, Neptune, Saturn , Jupiter, mars, Venus ,mercury and sun

4 0
2 years ago
A 4 kilogram box is at rest on a frictionless floor. A net force of 12 newtons acts on it. What is the acceleration of the box?
Rashid [163]

Answer:

<h2>3 m/s^2</h2>

Explanation:

Step one:

given

Mass m= 4kg

Force F= 12N

Required

Acceleration the relation between force, acceleration, and mass is Newton's first equation of motion, which says a body will continue to be at rest or uniform motion unless acted upon by an external force

F=ma

a=F/m

a=12/4

a=3 m/s^2

4 0
3 years ago
A proton moves from a location where V = 87 V to a spot where V = -40 V. (a) What is the change in the proton's kinetic energy?
Art [367]

Answer: a) 127 eV; b) there is no change of kinetic energy.

Explanation: In order to explain this problem we have to use the change of potentail energy ( conservative field) is equal to changes in kinetic energy. So for the proton ther move to lower potential then they gain kinetic energy from the electric field.  This means the electric force do work in this trayectory and then the protons increased changes its speed.

If we replace the proton by a electron we have a very different situaction, the electrons are located in a lower potental then  they can not move to higher potential  if any  external force does work on the system.

In resumem, the electrons do not move from a point with V=87 to other point with V=-40 V. The electric force point to high potential so the electrons  can not move to lower potential region (V=-40V).

6 0
3 years ago
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