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ruslelena [56]
3 years ago
8

A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.

What is the horizontal distance covered by the cannonball before it hits the ground?
Physics
1 answer:
pochemuha3 years ago
8 0

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t  = 6.55 \ s

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

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At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is is de
soldi70 [24.7K]

Answer:

2.9 N

Explanation:

When the separation distance, r, is 0.5 m, the electrostatic force is 0.32 N. Electrostatic force is given as:

F = (k * q1 * q2) / r²

Where F = force acting on the balloons

k = Coulombs constant

Therefore:

0.32 = (k * q1 * q2) / 0.5²

=> k * q1 * q2 = 0.32 * 0.5² ------------(1)

When the distance is decreased by 3, that is r = r/3 = 0.5/3

F = (k * q1 * q2) / (0.5/3)² ------------(2)

Putting (1) into (2):

=> F = (0.32 * 0.5²) / (0.5/3)²

F = (0.32 * 0.5² * 3²) / 0.5²

F = 2.9 N

Therefore, the force would be 2.9 N

3 0
3 years ago
which planet should punch travel to if his goal is to weigh in at 118 lb? refer to the table of planetary masses and radii given
Harrizon [31]

The planet that Punch should travel to in order to weigh 118 lb is Pentune.

<h3 /><h3 /><h3>The given parameters:</h3>
  • Weight of Punch on Earth = 236 lb
  • Desired weight = 118 lb

The mass of Punch will be constant in every planet;

W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}

where;

  • M is the mass of Earth = 5.972 x 10²⁴ kg
  • R is the Radius of Earth = 6,371 km

For Planet Tehar;

g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g

For planet Loput:

g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g

For planet Cremury:

g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g

For Planet Suven:

g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g

For Planet Pentune;

g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g

For Planet Rams;

g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g

The weight Punch on Each Planet at a constant mass is calculated as follows;

W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

<u>The </u><u>complete question</u><u> is below</u>:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).

<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>

Learn more about effect of gravity on weight here: brainly.com/question/3908593

5 0
2 years ago
If u shine a light of frequency 375Hz on a double slit setup, and u measure the slit separation to be 950nm and the screen dista
xxMikexx [17]

Answer:

Y = 3.39 x 10⁶ m

Explanation:

We will use Young's Double Slit formula here:

Y = \frac{\lambda L}{d}

where,

Y = Fringe spacing = ?

λ = wavelength = \frac{speed\ of\ light}{frequency} = \frac{3\ x\ 10^8\ m/s}{375\ Hz} = 8 x 10⁵ m

L = screend distance = 4030 nm = 4.03 x 10⁻⁶ m

d = slit separation = 950 nm = 9.5 x 10⁻⁷ m

Therefore,

Y = \frac{(8\ x\ 10^5\ m)(4.03\ x\ 10^{-6}\ m)}{9.5\ x\ 10^{-7}\ m}

<u>Y = 3.39 x 10⁶ m</u>

6 0
3 years ago
What the motion of an object that has an acceleration of 0 m/s
Degger [83]
<span>Everything in the system is stable and therefore the objects motion is stable. That is to say it is not changing what it is already doing. As far as i know zero times zero is still zero. In that case then the motion must be constant or stable.</span>
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the answer is A it is so easy question

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3 years ago
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