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AnnyKZ [126]
3 years ago
5

a ball of mass 0.5 kg is released from rest at a height of 30 m. how fast is it going when it hits the ground? acceleration due

to gravity is g = 9.8 m/s^2
Physics
1 answer:
Vera_Pavlovna [14]3 years ago
8 0

Answer:

24.25m/s

Explanation:

m = 0.5kg

u ( initial velocity) = 0m/s

v ( final velocity) =?

a = 9.8m/s^2

d (displacement) = 30m

Since u don't have time, u only have the choice to use this formula

V^2 = u^2 + 2ad

V^2 = 0 + 2 x 9.8 x 30

V^2 = 588

V = 24.25 m/s

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A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r <
Svet_ta [14]

Answer:  

Hence the answer is E inside = KQr_{1} /R^{3}.

Explanation:  

E inside = KQr_{1} /R^{3}  

so if r1 will be the same then  

E  \begin{bmatrix}Blank Equation\end{bmatrix} proportional to 1/R3  

so if R become 2R  

E becomes 1/8 of the initial electric field.

8 0
3 years ago
Read 2 more answers
Displacement vectors of 4km north, 2km south, 5km north, 5km south combine to a total displacement of
goldfiish [28.3K]

<u>Answer</u>

The combined displacement is 2km north


<u>Explanation</u>


Since displacement is a vector quantity, we take into account the direction.


Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.


We now add to obtain,

4+-2+5+-5

This will simplify to

=4-2+5-5=2

Therefore the combined displacement is 2km north

5 0
3 years ago
Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i
UkoKoshka [18]

Answer:

A. Zero

Explanation:

Given data,

The charge of the test charge, q = 1 C

The distance the charge moved against the filed of intensity, x = 30 cm

                                                                                                        = 0.3 m

The electric field intensity, E = 50 N/C

The energy stored in the charge at 0.3 m is given by the formula,

                                V = k q/r

Where,                        

                                     = 9 x 10⁹ Nm²C⁻²

The charge is moved from the potential V₁ to V₂ at 30 cm

Substituting the given values in the above equation

                            V₁ = 9 x 10⁹ x 30 / 0.3

                                =  1.5 x 10¹² J

And,

                            V₂ = 1.5 x 10¹² J

The energy stored in it is,

                             W = V₂ - V₁

                                  = 0

Hence, the energy stored in the charge is, W = 0        

6 0
3 years ago
7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the ca
Misha Larkins [42]

Answer:

q = C V    charge on 1 capacitor

q = 1 * 10E-6 * 110 = 1.1 *  10E-4  C per capacitor

N = Q / q = 1 / 1.1 * 10E-4  = 9091 capacitors

8 0
3 years ago
You approach a railroad crossing with a cross buck sign that has no lights or gates. What should you do?
maxonik [38]
Stop and look to see if anything is coming 
6 0
3 years ago
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