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yanalaym [24]
3 years ago
9

Can some help me pliz ​

Physics
1 answer:
muminat3 years ago
3 0

1).  a

2).  g

3).  b

4).  h

5).  e

6).  c

7).  d

8).  f

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A small object carrying a charge of -4.00 nC is acted upon by a downward force of 19.0 nN when placed at a certain point in an e
Gala2k [10]

Explanation:

Given that,

Charge acting on the object, q=-4\ nC=-4\times 10^{-9}\ C

Force acting on the object, F=19\ nC=19\times 10^{-9}\ C (in downward direction)

(a) The electric force acting in the electric field is given by :

F=qE

E is the electric field

E=\dfrac{F}{q}

E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, q=1.6\times 10^{-19}\ C

The force acting on the proton is :

F=qE

F=1.6\times 10^{-19}\times 4.75

F=7.6\times 10^{-19}\ N

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.

8 0
3 years ago
After hitting the spring, the block is bounced back up the ramp. The maximum compression of the spring is Δx=0.03m, and the spri
Lyrx [107]

Answer:

v = 1.28 m/s

Explanation:

Given that,

Maximum compression of the spring, \Delta x=0.03\ m

Spring constant, k = 800 N/m

Mass of the block, m = 0.2 kg

To find,

The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.

Solution,

When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,

Initial energy = Final energy

\dfrac{1}{2}kx^2=mgh+\dfrac{1}{2}mv^2

Substituting all the values in above equation,

\dfrac{1}{2}\times 800\times 0.03^2=0.2\times 9.8\times 0.1+\dfrac{1}{2}\times 0.2\times v^2

v = 1.28 m/s

Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.

5 0
2 years ago
Help me plssssssss cause I’m struggling
krok68 [10]

Answer:

I am pretty sure it is C

Explanation:

It can be found all over the universe

5 0
3 years ago
Read 2 more answers
A stone is thrown vertically upward with a speed of 18.0 . (a)How fast is it moving when it reaches a height of 11.0 ? (b)How lo
aliina [53]
For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative. 
So <span>Vf = square root [2(-9.8)(11.0) + (18.0)^2] </span>
<span>Vf = 10.4 m/s your answer is correct. 

For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, </span>11= 18t m/s-(1/2) 9.8t^2 then <span>-11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s</span><span>
 </span> 
5 0
3 years ago
What is the order of magnitude of the distance of Sun to nearest star in meters?
neonofarm [45]

Answer:

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:

V

=

π

⋅

r

2

⋅

h

Plugging in our numbers (and assuming that

π

≈

3

)

V

=

π

⋅

(

10

21

m

)

2

⋅

(

10

19

m

)

V

=

3

×

10

61

m

3

Is the approximate volume of the Milky Way.

Now, all we need to do is find how many stars per cubic meter (

ρ

) are in the Milky Way and we can find the total number of stars.

Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of

4

×

10

16

m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.

ρ

=

n

V

Using the volume of a sphere

V

=

4

3

π

r

3

ρ

=

1

4

3

π

(

4

×

10

16

m

)

3

ρ

=

1

256

10

−

48

stars /

m

3

Going back to the density equation:

ρ

=

n

V

n

=

ρ

V

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

n

=

(

1

256

10

−

48

m

−

3

)

⋅

(

3

×

10

61

m

3

)

n

=

3

256

10

13

n

=

1

×

10

11

stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.

4 0
2 years ago
Read 2 more answers
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