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3 years ago

8

Astronauts need special tools to work in space. One specially designed tool that astronauts use is a screwdriver. Using Newton’s

third law of motion, explain why it might be difficult to use a normal screwdriver to tighten a screw in space.

1 answer:

Softa [21]3 years ago

3 0

Due to the fact that for every action there is an equal reaction for every turn on the screw the screw will exert energy equal to the amount you gave

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Velocity vs Time

Grace [21]

Answer:

A and C

C

40 m/s

zero

Explanation:

8 0

3 years ago

A body of mass m1 = 1.5 kg moving along a directed axis in the positive sense with a velocity

larisa86 [58]

Answer:

3.71 m/s in the negative direction

Explanation:

From collisions in momentum, we can establish the formula required here which is;

m1•u1 + m2•v2 = m1•v1 + m2•v2

Now, we are given;

m1 = 1.5 kg

m2 = 14 kg

u1 = 11 m/s

v1 = -1 m/s (negative due to the negative direction it is approaching)

u2 = -5 m/s (negative due to the negative direction it is moving)

Thus;

(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)

This gives;

16.5 - 70 = -1.5 + 14v2

Rearranging, we have;

16.5 + 1.5 - 70 = 14v2

-52 = 14v2

v2 = - 52/14

v2 = 3.71 m/s in the negative direction

8 0

3 years ago

Which image represents the force on a positively charged particle caused by an approaching magnet?

Amanda [17]

Answer:

Image B represents the force on a positively charged particle caused by an approaching magnet.

Explanation:

The most fundamental law of magnetism is that like shafts repulse each other and dissimilar to posts pull in one another; this can without much of a stretch be seen by endeavoring to put like posts of two magnets together. Further attractive impacts additionally exist. On the off chance that a bar magnet is cut into two pieces, the pieces become singular magnets with inverse shafts. Also, pounding, warming or winding of the magnets can demagnetize them, on the grounds that such dealing with separates the direct game plan of the particles. A last law of magnetism alludes to maintenance; a long bar magnet will hold its magnetism longer than a short bar magnet. The domain theory of magnetism expresses that every single enormous magnet involve littler attractive districts, or domains. The attractive character of domains originates from the nearness of significantly littler units, called dipoles. Iotas are masterminded in such a manner in many materials that the attractive direction of one electron counteracts the direction of another; in any case, ferromagnetic substances, for example, iron are unique. The nuclear cosmetics of these substances is with the end goal that littler gatherings of particles unite as one into zones called domains; in these, all the electrons have the equivalent attractive direction.

4 0

3 years ago

Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

5 0

4 years ago

A solar panel measures 0.57 m by 1.3 m. In direct sunlight, the panel delivers 6.6 A at 6 V. If the intensity of sunlight is 300

zheka24 [161]

Answer:

efficiency of solar panel is 18%

Explanation:

Energy received by the sunlight is given as

$Power = intensity \times area$

here we know the dimensions of the plate as

$L = 0.57 m$

$W = 1.3 m$

now we have

$Area = (0.57)(1.3) m^2$

$A = 0.741 m^2$

now the power received by the sun light is given as

$P_1 = 0.741(300) = 222.3 W$

now the output power due to solar panel is given as

$P_{out} = V i$

$P_{out} = (6.6 A)(6 V)$

$P_{out} = 39.6 Watt$

now the efficiency is given as

$\eta = \frac{P_{out}}{P_{in}}$

$\eta = \frac{39.6}{222.3}$

$\eta = 0.18$

8 0

4 years ago