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Scrat [10]
3 years ago
14

PLEASE HELP ME! THIS IS TIMED

Physics
2 answers:
vova2212 [387]3 years ago
6 0
Vas happenin!!!

Hope your day is good

I would say observational tool

Hope this helps

*smiles*
siniylev [52]3 years ago
5 0

Answer:

C

Explanation:

Which type of tool was modified to view outer space? Observational tool.

(1) Quiz 4: Tools and Technology Flashcards

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11.
jolli1 [7]
<h3>Reducing Surface Area.</h3>

If the surface area becomes smaller, the pressure becomes larger.

8 0
3 years ago
You push on a 30 kg box with a force of 120 N. What is the acceleration of the box2​
kumpel [21]

Answer:

6

Explanation:

120/30=6

8 0
3 years ago
A car travels 50 km in 25 km /h and then travels 60km with a velocity 20 km/h and then travels 60km with a velocity 20 km/h in t
Butoxors [25]

Answer:

v = 21.25 km/h

The average velocity is 21.25km/h

Explanation:

Average velocity = total displacement/time taken

v = d/t

Given;

A car travels 50 km in 25 km /h

d1 = 50km

v1 = 25km/h

time taken = distance/velocity

t1 = d1/v1

t1 = 50/25 = 2 hours

and then travels 60km with a velocity 20 km/h

d2 = 60km

v2 = 20km/h

t2 = d2/v2 = 60/20

t2 = 3 hours

and then travels 60km with a velocity 20 km/h in the same direction

d3 = 60km

v3 = 20km/h

t3 = d3/v3 = 60/20

t3 = 3 hours

Average velocity = total displacement/total time taken

v = (d1+d2+d3)/(t1+t2+t3)

v = (50+60+60)/(2+3+3)

v = 170/8

v = 21.25 km/h

The average velocity is 21.25km/h

3 0
3 years ago
what is the difference between kinetic energy and random motion of a cylider of oxygen carried by a car and one standing on a pl
pochemuha

Answer:

See explanation below

Explanation:

If we are talking about the kinetic energy of the cylinder of oxygen:

The kinetic energy possessed by any object is given by

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

where

m is the mass of the object

v is its speed

In this case, we have one cylinder carried by a car and one standing on a platform: this means that the speed of the cylinder carried by the car will be different from zero (and so also its kinetic energy will be different from zer), while the speed of the cylinder standing on the platform will be zero (and so its kinetic energy also zero). Therefore, the kinetic energy of the cylinder carried by the car will be larger than that standing on a platform.

Instead, if we are talking about the kinetic energy due to the random motion of the molecules of oxygen inside the cylinder:

The kinetic energy of the molecules in a gas is directly proportional to the absolute temperature of the gas:

K=\frac{3}{2}kT

where k is called Boltzmann constant and T is the absolute temperature of the gas. Therefore, we see that K does not depend on whether the gas is in motion or not, but only on its temperature - therefore, in this case there is no difference between the kinetic energy of the cylinder carried by the car and that standing on the platform (assuming they are at the same temperature)

6 0
3 years ago
A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes fro
Amanda [17]

Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

(B). We need to calculate the magnetic flux

Using formula of flux

\phi_{1}=BA\cos\theta

Put the value into the formula

\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0

\phi_{1}=0.01021\ Wb

We need to calculate the magnetic flux

Using formula of flux

\phi_{2}=BA\cos\theta

Put the value into the formula

\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0

\phi_{2}=0.00127\ Wb

We need to calculate the magnitude of the average induced emf

Using formula of emf

\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})

Put the value into t5he formula

\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})

\epsilon=0.016867\ V

\epsilon=16.87\ mV

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

I=\dfrac{\epsilon}{R}

Put the value into the formula

I=\dfrac{0.016867}{2.5}

I=0.00675\ A

I=6.75\ mA

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

5 0
3 years ago
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