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iren2701 [21]
3 years ago
6

A brick is thrown vertically upward with an initial speed of 3.00 m/s from the roof of a building. If the building is 78.4 m tal

l, how much time passes before the brick lands on the ground?
Physics
2 answers:
masya89 [10]3 years ago
6 0

Answer:

0.918 sec

Explanation:

time of flight:

t=u²/g

=3²/9.8=0.918sec

natali 33 [55]3 years ago
3 0

Answer:

4.32 seconds

Explanation:

Given:

y = 0 m

y₀ = 78.4 m

v₀ = 3.00 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 = 78.4 + 3.00 t − 4.9 t²

4.9t² − 3t − 78.4 = 0

Solve with quadratic formula:

t = [ 3 ± √(9 − 4(4.9)(-78.4)) ] / 9.8

t ≈ -3.71, 4.32

Since t must be positive, t = 4.32.  So it takes 4.32 seconds for the brick to land.

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<h2><u>Hello </u><u>There</u></h2>

\tt{Difference \:  Between \:   \underline\red {Horticulture} \:  And \:   \underline\green{Agriculture}}

\text{\underline\red {Horticulture}}

The Cultivation of Fruits, Vegetables and Flowers for domestic and international markets are called Horticulture.

\text{\underline\green{Agriculture}}

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2 years ago
The drawing shows a model for the motion of the human forearm in throwing a dart. Because of the force M applied by the triceps
Serga [27]

Answer:

464.3 N

Explanation:

Given parameters are:

I = 0.065 kg*m^2

L = 0.025 m

R = 0.28 m

v_0 = 0 m/s

v_f = 5 m/s

t = 0.1 s

v_f=v_0+at=at

Hence, a=v_f/t

We must connect two torque equations to find the answer.

\tau=LM=I\alpha

Where \alpha =\frac{a}{R} =\frac{v_f}{Rt}

Hence, LM=I\frac{v_f}{Rt}

Thus, M = \frac{Iv_f}{LRt} = \frac{0.065*5}{0.025*0.28*0.1} =464.3 N

5 0
4 years ago
Describe the role of observations in scientific endeavor and explain the characteristics of good observations.
koban [17]
Observations are used in order to collect data and record a variety of interesting or useful key points about the subject or specimen in observation. These observations, if made well, can be recorded and used to supplement a hypothesis. 
3 0
3 years ago
A snowball starting at rest rolls down a hill and reaches 5 m/s. If the hill is
Lostsunrise [7]

Answer:

The acceleration of the snowball is 0.3125

Explanation:

The initial speed of the snowball up the hill, u = 0

The speed the snowball reaches, v = 5 m/s

The length of the hill, s = 40 m

The equation of motion of the snowball given the above parameters is therefore;

v² = u² + 2·a·s

Where;

a = The acceleration of the snowball

Plugging in the values, we have;

5² = 0² + 2 × a × 40

∴ 2 ×  40 × a  = 5² = 25

80 × a = 25

a = 25/80 = 5/16

a = The acceleration of the snowball = 5/16 m/s².

The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .

4 0
3 years ago
A bowling ball with a mass of 7.0kg strikes a pin that had a mass of 2.0kg the pin flies forward with a velocity of 6.0m/s, and
Natalka [10]

The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.

We don't have external forces, so:

P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}\\

Where:

  • mb is the mass of the bowling ball
  • mp the mass of the pin
  • v_{0b}\quad and\quad v_{0p} the initial velocities of the bowling ball and the pin.
  • v_{1b}\quad and\quad v_{1p} the final velocities of the bowling ball and the pin.

Solving for v0b:

v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}

<h2>R/ The original velocity of the ball was 5.71 m/s.</h2>
6 0
3 years ago
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