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stellarik [79]
3 years ago
11

How would I graph the data in the table on the graph? ********WORTH 50 POINTS********WILL MARK BRAINLIEST**********

Physics
2 answers:
Soloha48 [4]3 years ago
5 0

Answer:

you count to the right (time) and you count up in distant and put a dot . BOOM  

Explanation:

Natali5045456 [20]3 years ago
3 0

Answer:

Save the file and try to put stuff over the file. (kinda off like a picture)

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A boy pushes his baby sister in a stroller that has a mass of 45,000 g. What is the acceleration on the stroller if the force is
dlinn [17]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Force.

Force = Mass * Acceleration.

here, Mass = 45 kg and Force = 65 N

hence, Acceleration = 65/45

===> Acc. = 1.44 m/s^2

hence the acceleration is 1.44 m/s^2

8 0
3 years ago
How are defenders on a baseball team positioned ?
Ulleksa [173]

Answer:

1 (pitcher), 2 (catcher), 3 (first baseman), 4 (second baseman), 5 (third baseman), 6 (shortstop), 7 (left fielder) 8 (center fielder), and 9 (right fielder)

Explanation:

There are nine fielding positions in baseball. Each position conventionally has an associated number, for use in scorekeeping by the official score

8 0
3 years ago
Read 2 more answers
A lamp can work on 50V mains taking 2 amps. What value of resistance must be connected in series with it, so that it can be oper
wolverine [178]
The resistance of the lamp is apparently  50V/2A  =  25 ohms.

When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.

In order for 200V to cause 2A of current, the total resistance
must be      200V/2A = 100 ohms.

The lamp provides 25 ohms, so we want to add another 75 ohms 
in series with the lamp.  Then the total resistance of the circuit is
(75 + 25) = 100 ohms, and the current is 200V/100 ohms = 2 Amps.

The power delivered by the 200V mains is (200V) x (2A) = 400 watts.

The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.

The extra resistor dissipates  ( I² · R) = (2² · 75 ohms) = 300 watts.

Together, they add up to the 400 watts delivered by the mains.

CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it. 
It has to be a special 'power resistor'. 
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms.  Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.
3 0
3 years ago
A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. I
vodka [1.7K]

Answer:

a = 603.59 m/s^2

Explanation:

from the data given . the rate of change in magnetic field is as follow

\frac{dB}{dt} = 280 G/s = 280 \times 10^{-4} T/s

from the faraday's law of induction , the expression for the induced emf in region of radius r as follow

\epsilon = \frac{d \phi}{dt}

\int E.dl = \frac{d(BA)}{dt}

E(2\pi r)= \pi r^2 \frac{dB}{dt}

E = \frac{r}{2} \frac{dB}{dt}

electric field at point P_1 as follow

E = \frac{r}{2} \frac{dB}{dt}

E = \frac{1.5\times 10^{-2}}{2} 280 \times 10^{-4}

E = 6.3\times 10^{-6} V/m

from newton 2nd law of motion, the acceleration of proton is

F = ma

qE = ma

a = \frac{qE}{m}

a = \frac{1.6 \times 10^{-19} (6.3\times 10^{-6})}{1.67\times 10^{-27}}

a = 603.59 m/s^2

5 0
3 years ago
When the temperature of matter decrease , the particles do what
Rom4ik [11]

Answer:

When the temperature decreases the particals start to slow down.

4 0
3 years ago
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