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GuDViN [60]
3 years ago
9

Listed in the Item Bank are some important labels for sections of the image below. To find out more information about labels, so

me have more details available when you click on them. Drag and drop each label to the corresponding area it identifies in the image.

Physics
1 answer:
erica [24]3 years ago
5 0

1 - cytoplasm

2 - DNA

3 - rybosome

4 - cell membrane

5 - nucleus

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An object of mass 100kg is released from rest and falls through a distance of 10m.what is the work done by gravity
aleksandr82 [10.1K]

Answer:

Mass = 100 kg

Acceleration due to gravity = 9.8m/s

Height = 10 m

Work done = mgh

= 100 x 9.8 x 10

= 9800 J

Hope this helps!

8 0
3 years ago
A block is on a frictionless table, on earth. The block accelerates at 7.5 m/s when a 70 N horizontal force is applied to it. Th
Liono4ka [1.6K]

Answer:

The weight of the block on the moon is 15 kg.

Explanation:

It is given that,

The acceleration of the block, a = 7.5 m/s²

Force applied to the box, F = 70 N

The mass of the block will be, m=\dfrac{F}{a}

m=\dfrac{70\ N}{7.5\ m/s^2}

m = 9.34 kg

The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :

W=m\times g

W=9.34\ kg\times 1.62\ m/s^2

W = 15.13 N

or

W = 15 N

So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.

3 0
3 years ago
Is it ok as a headline story
katen-ka-za [31]
Ya it looks great to be in a story
8 0
3 years ago
Read 2 more answers
Which is an example of something heated by conduction?
babunello [35]

Answer:

A waffle iron heated by coils

Explanation:

A waffle iron heated by coils - conduction

Food heated in a microwave oven - radiation

Pavement heated by the sun - radiation

A room heated by moving air - convection

5 0
3 years ago
Read 2 more answers
A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

F_g = F_c

\frac{GmM}{r^2} = \frac{mv^2}{r}

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0
3 years ago
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