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kaheart [24]
3 years ago
12

A soccer ball is kicked horizontally off a bridge with a heigh of 36m. The ball travels 25m horizontally before it hits the pave

ment below. What was the soccer ball's speed when it was first kicked?
Physics
1 answer:
kow [346]3 years ago
5 0

Answer:

The initial velocity was U=22.14m/s

Explanation:

Step one :

Applying the third equation of motion

v² = u²+ 2as

Where v= Final velocity

U =initial velocity

a= acceleration due to gravity

S= distance or displacement

Step two :

V= 0

a= 9.81m/s²

S=25m

U=?

Step three :

Substituting into the equation we have

0²=U²+2*9.81*25

0=U²+490.5

U²=-490.5

U=√490.5

U=22.14m/s

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When work is done on a spring to stretch it, elastic potential energy is stored in the spring.
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Explanation:

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Two identical point charges are 3.00 cm apart. find the charge on each of them if the force or repulsion is 4.00 x 10^-7. (Use C
DanielleElmas [232]

Answer:

Charge on each is 2 x 10⁻¹⁰.

Explanation:

We know that Force between two point charges is given b the Coulomb's law as:

F = kq₁q₂/r^2

k = 9 x 10^9

r = 3.00 cm

= 0.03 m

q₁ = q₂

F = 4.00 x 10^-7

Rearranging the formula, we get:

F = k q²/r²

q² = Fr²/k

q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)

q² = 4 x 10⁻²⁰

q = 2 x 10⁻¹⁰

As there is force of repulsion between the charges, the charges must be both positive or both negative.

3 0
3 years ago
Options: A.) 10 N B.) 15 N C.) 25 N D.) 35N
Lady_Fox [76]

Given:

F_gravity = 10 N

F_tension = 25 N

Let's find the net centripetal force exterted on the ball.

Apply the formula:

\sum ^{}_{}F_{\text{net}}=F_1+F_2=F_{centripetal}

From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.

Hence, the tensional force is positive while the gravitational force is negative.

Thus, we have:

F_{\text{net}}=F_{\text{centripetal}}=F_{tension}-F_{gravity}=25N-10N=15N

Therefore, the net centripetal force exterted on the ball is 15 N.

ANSWER:

15 N

7 0
1 year ago
Four charges are on the four corners of a square. Q1 = +5μC, Q2 = -10μC, Q3 = +5μC, Q4 = -10μC. The side length of the square is
Marat540 [252]

Answer:

Explanation:

Electric field due to a point charge Q at a point at distance d is given by the relation

E = \frac{K\times Q}{d^2}

Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.

The charges are situated on the corners of a square in such a way that

equal charges of Q1 and Q3 are situated on the diametrically  opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two  charges will be zero.  

On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero

Overall, net field due to all the four charges will be zero

3 0
3 years ago
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