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ira [324]
2 years ago
12

Pls help me with this question its due tomorrow and I can't find the answer: A person jumped 50cm, the hang time for the jump wa

s 0.778 seconds, calculate a value for the acceleration due to gravity. What is the percent error in the value for acceleration you found? (Solve it without factoring in the air)

Physics
1 answer:
son4ous [18]2 years ago
4 0

S = ut + 1/2gt²

where S = distance

50 cm = 0.5m

u is initial velocity = 0

S = 0*t + 1/2gt²

S = 1/2gt²

0.5 = 1/2 g (0.778²)

2*0.5 = g(0.778²)

1/(0.778²) = g

g = 1.65m/s²

The value of g is grossly errored due to the values obtained since the known value of g on earth is 9.8m/s² the obtained value rounds off to the value of g on moon.

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A 10.0 N package of whole wheat flour is suddenly placed on the pan of a scale such as you find in grocery stores. The pan is su
inna [77]

Answer:

0.025 m

Explanation:

From the question,

Applying Hook's law

F = ke................... Equation 1

Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.

make e the subject of the equation

e = F/k....................... Equation 2

Given: F = 10 N, e = 395 N/m

Substitute these values into equation 2

e = 10/395

e = 0.025 m

7 0
3 years ago
a student pushes on a crate with a force of 100 N directed to the right. what force does the crate exert on a student
soldi70 [24.7K]

100N to the left. Newton's 3rd law action and reaction

5 0
2 years ago
Two identical freight cars roll without friction (one at 1 m/s, the other at 2 m/s) toward each other on a level track. They col
Kazeer [188]

Answer:

They collide, couple together, and roll away in the direction that <u>the 2m/s car was rolling in.</u>

Explanation:

We should start off with stating that the conservation of momentum is used here.

Momentum = mass * speed

Since, mass of both freight cars is the same, the speed determines which has more momentum.

Thus, the momentum of the 2 m/s freight car is twice that of the 1 m/s freight car.

The final speed is calculated as below:

mass * (velocity of first freight car) + mass * (velocity of second freight car) = (mass of both freight cars) * final velocity

(m * V1) + (m * V2) = (2m * V)

Let's substitute the velocities 1m/s for the first car, and - 2m/s for the second. (since the second is opposite in direction)

We get:

m*1 + m*(-2) = 2m*V

solving this we get:

V = - 0.5 m/s

Thus we can see that both cars will roll away in the direction that the 2 m/s car was going in. (because of the negative sign in the answer)

7 0
3 years ago
a motorcycle accelerates from 15 m/s to 20 m/s over a distance of 50 meters. what is its average acceleration?
devlian [24]
For this, you need the v-squared equation, which is v(final)² = v(initial)² + 2aΔx The averate acceleration is thus a = (v(final)² - v(initial)²) / 2Δx = (20² - 15²) / 2(50) = 175 / 100 = 1.75 m/s² So the average acceleration is 1.75 m/s²
5 0
3 years ago
Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0
2 years ago
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