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3 years ago

A seed can be round when parents are crossed, scientists refer to the first generation of offspring as

1 answer:

4 0

Hey there!

A seed can be round when parents are crossed, scientists refer to the first generation of offspring as F1. This is the first generation of seeds from very different parental types.

Thank you!

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A 59.31 kg rock is sitting at the top of a cliff that is 300 m tall. What is the gravitational potential energy of that rock?

Diano4ka-milaya [45]

Answer:

The gravitational potential energy of that rock is 174371.4 J.

Explanation:

Given

Mass m = 59.31 kg

Height h = 300 m

To determine

We need to find the gravitational potential energy of the rock

We know that the potential energy of a body is termed as the stored energy due to its position.

One kind of energy comes from Earth's gravity — Gravitational potential energy (GPE).

Gravitational potential energy (GPE) can be determined using the formula

$GPE = mgh$

where

$m$ is the mass

$g$ is the gravitational acceleration which is equal to g = 9.8 m/s²

$h$ is the height

GPE is the Gravitational potential energy

now substituting m = 59.31, h = 300 and g = 9.8

$GPE = mgh$

$=59.31\times 9.8\times 300$

$=174371.4$ J

Therefore, the gravitational potential energy of that rock is 174371.4 J.

4 0

3 years ago

The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the

xz_007 [3.2K]

Answer:

a) -1.14 rev/min²

b) 9900 rev

c) -9.92×10⁻⁴ m/s²

d) 30.8 m/s²

Explanation:

First, convert hours to minutes:

2.2 h × 60 min/h = 132 min

a) Angular acceleration is change in angular velocity over change in time.

α = (ω − ω₀) / t

α = (0 rev/min − 150 rev/min) / 132 min

α = -1.14 rev/min²

b) θ = θ₀ + ω₀ t + ½ αt²

θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²

θ = 9900 rev

c) The tangential component of linear acceleration is:

a_t = αr

First, convert α from rev/min² to rad/s²:

-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²

Therefore:

a_t = (-1.98×10⁻³ rad/s²) (0.50 m)

a_t = -9.92×10⁻⁴ m/s²

d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:

a² = (a_t)² + (a_r)²

The radial component is the centripetal acceleration:

a_r = v² / r

a_r = ω² r

First, convert 75 rev/min to rad/s:

75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s

Find the radial component:

a_r = (7.85 rad/s)² (0.50 m)

a_r = 30.8 m/s²

Now find the net linear acceleration:

a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²

a = 30.8 m/s²

5 0

3 years ago

A person run 21.0 km west then runs 10.0 km east what is the persons distance

VashaNatasha [74]

Answer:

31.0 km

Explanation:

21.0+10.0=31.0 km

6 0

3 years ago

Read 2 more answers

When this circuit is closed, which way do the electrons flow?

gizmo_the_mogwai [7]

Answer:

from the positive end of the battery through the capacitor through the resistors to the negative end...

Current flows from higher potential (+) to lower potential (-)..

3 0

3 years ago

Read 2 more answers

A bullet is fired straight up from a gun with a

Katena32 [7]

v(4 seconds)= 300 m/s - 9.8 (m/s^2)(4s) = 260.8 m/s , hope this helps:)

6 0

3 years ago

Read 2 more answers