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Luden [163]
3 years ago
8

A student releases a ball from a height of 100 cm above the ground. It bounces to a height of 80 cm. Answer the following questi

ons according to the diagram

Physics
1 answer:
Zepler [3.9K]3 years ago
4 0

if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...

if α = 45°, then the equation may be written as: ...

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.

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Which of the following defines the Permian Triassic​
Mama L [17]

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The Permian–Triassic extinction event, also known as the P–Tr extinction, the P–T extinction, the End-Permian Extinction and colloquially as the Great Dying, formed the boundary between the Permian and Triassic geologic periods, as well as between the Paleozoic and Mesozoic eras, approximately 252 million years ago

8 0
3 years ago
What type of Circuit do they use in a home? WHY ? (Simple circuit / Parallel circuit / Series circuit) helpp
Vikentia [17]

Answer:

Parallel

Explanation:

Most standard 120-volt household circuits in your home are (or should be) parallel circuits. Outlets, switches, and light fixtures are wired in such a way that the hot and neutral wires maintain a continuous circuit pathway independent from the individual devices that draw their power from the circuit.

Hope that helps :)

7 0
3 years ago
Be sure to answer all parts. Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium
ss7ja [257]

Answer:

The number of atoms are 1.86\times10^{8}.

Explanation:

Given that,

Diameter D = 1.40\times10^{2}\ pm

D=1.40\times10^{2}\times10^{-12}\ m

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}

Number\of\atoms =1.86\times10^{8}

Hence, The number of atoms are 1.86\times10^{8}.

8 0
3 years ago
Can someone please help me with this physics question? I'm desperate!
Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

3 0
3 years ago
Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 s to speed up from rest t
Vinvika [58]

Answer

Time period T = 1.50 s

time t = 40 s

r = 6.2 m

a)

Angular speed ω = 2π/T

                              = \dfrac{2\pi }{1.5}  

                              = 4.189 rad/s

Angular acceleration α = \dfrac{\omega}{t}

                                      = \dfrac{4.189}{40}

                                      = 0.105 rad/s²

Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²

b)The maximum speed.

       v = 2πr/T

          = \dfrac{2\pi \times 6.2}{1.5}

          = 25.97 m/s

So centripetal acceleration.

        a = \dfrac{v^2}{r}

          = \dfrac{25.97^2}{6.2}

          =  108.781 m/s^2

          = 11.1 g    

in combination with the gravitation acceleration.

a_{total} = \sqrt{(11.1g)^2+g^2}

a_{total}= 11.145 g

6 0
4 years ago
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