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3 years ago

_____ help by supporting the plant and holding the leaves up to the light.

Physics

2 answers:

Semenov [28]3 years ago

7 0

The answer is the stem.

pochemuha3 years ago

3 0

Answer:

The Answer is Stem :D

Explanation:

Because the stem is the part that holds the plant to the light.

Hope i helped

and again the answer is the stem.. :D

Have a great day!!

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A machine lifts a 35 kg object doing 6860 J worth of work. How much power is produced by the machine if it lifts the object in 4

timofeeve [1]

Answer:

Explanation:

We need the power equation for this which is

P = Work/time

We have everything we need to solve this (the mass of the object is extra information):

P = 6860/4

P = 1715W

5 0

2 years ago

A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe

Sindrei [870]

Answer:

The magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ .

Explanation:

Given that,

Charge, $q=-8.5\ \mu C=-8.5\times 10^{-6}\ C$

Speed of the charged particle, $v=9\times 10^6\ m/s$

The angle between the velocity of the charge and the field is 56°.

The magnitude of force, $F=5.9\times 10^{-3}\ N$

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

$F=qvB\ \sin\theta$

B is the magnetic field.

$B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T$

So, the magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ . Hence, this is the required solution.

4 0

3 years ago

A bird lands on a bare copper wire carrying a current of 51

nirvana33 [79]

The resistance of the piece of wire is
$R= \frac{\rho L}{A}$
where
$\rho = 1.68 \cdot 10^{-8}\Omega m$ is the resistivity of the copper
$L=5.1 cm=0.051 m$ is the length of the piece of wire
$A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2$ is the cross sectional area of the wire
By substituting these values, we find the value of R:
$R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega$

Then, by using Ohm's law, we find the potential difference between the two points of the wire:
$V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V$

7 0

3 years ago

Jane, looking for tarzan, is running at top speed (5.2 m/s) and grabs a vine hanging 4.5 m vertically from a tall tree in the ju

kumpel [21]

While Jane is running has a kinetic energy, which is Ek = 1/2*m*v^2 where m is mass and v is velocity  When she grabs a vine, she is going to change the kinetic energy to potential energy.
We know that potential energy is given by Ep = m*g*h where m is mass, g is gravity constant and h is height
So while running the kinetic energy is Ek = 1/2 * m * 5.2^2 = 13.52*m
Then all that energy is used to swing upward and gain potential energy
Ep = m*g*h = Ek = 13.52*m
m*9.8*h = 13.52*m
h = 13.52/9.8 = 1.38 meters
So Jane will swing 1.38 meters upward

8 0

2 years ago

A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th

Valentin [98]

Answer:

The escape speed for the craft is 1.49 m/s.

Explanation:

In this case we need to find the escape speed for a craft launched from a space elevator at a height of 56,000 km. The escape velocity is given by :

$v=\sqrt{\dfrac{2GM}{d}}$

Here,

G is universal gravitational constant

M is mass of earth

d = r + h, r is radius of Earth

$v=\sqrt{\dfrac{2\times 6.67\times 10^{-11}\times 5.972 \times 10^{24}}{(6,371 \times 10^3\times 56000\times 10^3)}}\\\\v=1.49\ m/s$

So, the escape speed for the craft is 1.49 m/s.

0 0

3 years ago