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Anarel [89]
3 years ago
5

A lander vehicle on a distant planet records the fact that it takes 2.86s for a ball to fall freely 11.66m, starting from rest.

What is the acceleration due to gravity on that planet? Express your answer in m/s^2. I am getting -1.425, but according to the hw, the answer is wrong :/.
Physics
1 answer:
solong [7]3 years ago
3 0
Based on your question that ask for the acceleration due to gravity on a certain planet and to answer that you must first use this formula S=1/2AT^2 and the acceleration of that certain planet is 2.85m/s^2
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You are shooting at a moving target. you estimate the length of lead necessary to hit the target. you maintain that lead as you
uysha [10]

The answer to this problem is:

This method of shooting is called <u>“sustained lead”.</u>

<span>This method of shooting at moving targets requires a bit of experience. Since the accuracy of hitting the target relies on the estimated amount of lead, this is useful when the path and speed of the target is steady and known.</span>

5 0
3 years ago
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What is the person’s displacement for the trip?<br><br> what is a vector quantity? give an example.
borishaifa [10]

Answer:

+ 3.0 m

Explanation:

displacement is shortest distance from fixed point O in particular direction . in diagram shortest distance at end from O is 3 m and it is right of O so +. HENCE +3.0m

4 0
3 years ago
A 242-g block is pressed against a spring of force constant 1.62 kN/m until the block compresses the spring 10.0 cm. The spring
hammer [34]

Answer:

a) = 3.94 m

b) = 3.15 m

Explanation:

Given

Mass of the block, m = 242 g

Force constant, k = 1.62 kN/m

Compression of the spring, x = 10 cm

Angle of inclination = 60°

a) if we equate the energy at the bottom of the ramp to the energy at a distance d up the ramp, we have

1/2kx² = mgh where, h = dsinΦ

1/2kx² = mgdsinΦ

1/2 * 1.62*10^3 * 0.1² = 0.242 * 9.8 * dsin 60

1/2 * 16.2 = 2.3716 * d sin 60

d sin 60 = 8.1 / 2.3716

0.866 d = 3.415

d = 3.415 / 0.866

d = 3.94 m

b) net force on the block = mgd sin 60 + µ mgd cos 60

8.1 = d[mg sin 60 + µ mg cos 60]

8.1 = d [0.242 * 9.8 * 0.866 + 0.44 * 0.242 * 9.8 * 0.5]

8.1 = d (2.05 + 0.52)

8.1 = 2.57 d

d = 8.1 / 2.57

d = 3.15 m

3 0
3 years ago
Read 2 more answers
We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes
Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

1.56421\times 10^{-6}\ T

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = \frac{d}{2}=\frac{7}{2}=3.5\ cm

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

c = Speed of light = 3\times 10^8\ m/s

The intensity is given by

I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2

5% of this energy goes to the visible light so the intensity is

I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

u=\frac{1}{2}\epsilon_0E^2

Energy density is also given by

\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T

The amplitude of the magnetic field at this surface is 1.56421\times 10^{-6}\ T

7 0
3 years ago
Elle is playing with a ball in a bus that moves in a straight line with constant velocity. What can you say about the motion of
kramer
Given that they are all on the same bus that is travelling in a straight line at the same velocity, when Elle throws the ball directly upwards, the ball will simply fall back to her. This is because the bus, Elle, and the ball are all travelling in the same direction and at the same speed. Among the choices, the correct answer is A.
8 0
3 years ago
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