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uysha [10]
2 years ago
12

A soccer ball is kicked horizontally. What is the average speed if its distance is 21.0 m after 4.00 s?

Physics
1 answer:
Over [174]2 years ago
5 0
The answer would be 5.25 m/s
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A truck of mass 1800kg is moving with a speed 54km/h. When brakes are applied, it
EleoNora [17]

Force = 3200 N

Work done = 640, 000 Nm

Explanation:

We begin by calculating the deceleration of the truck, using the velocity and distance;

a = (v² – u²)/2s

whereby;

a = acceleration

v = initial velocity

u = initial velocity

s = distance

We begin by changing the speed from km/h into m/s;

54km/hr = 15m/s

Then acceleration;

a = (0² – 15²) / 2 * 200

a = -225 / 400

a = - 0.5625 m/ s²

To calculate force;

F = ma

Whereby;

F = force

M = mass (in kgs)

a = acceleration

F = 1800 / 0.5625

F = 3200 N

Work done = Force * displacement

Work done = 3200 * 200

= 640, 000 Nm

Learn More:

For more on force and work done check out;

brainly.com/question/8662583

brainly.com/question/1268612

brainly.com/question/11870590

brainly.com/question/9125094

#LearnWithBrainly

6 0
3 years ago
As long the sun will be in the red giant? Question about time period, not when
steposvetlana [31]
It will be a red giant for about 5 billion years
6 0
3 years ago
A 1.30-kg object is held 1.10 m above a relaxed, massless vertical spring with a force constant of 315 N/m. The object is droppe
pshichka [43]

Answer:

0.345m

Explanation:

Let x (m) be the length that the spring is compress. If we take the point where the spring is compressed as a reference point, then the distance from that point to point where the ball is held is x + 1.1 m.

And so the potential energy of the object at the held point is:

E_p = mgh

where m = 1.3 kg is the object mass, g = 10m/s2 is the gravitational acceleration and h = x + 1.1 m is the height of the object with respect to the reference point

E_p = 1.3 * 10 * (x + 1.1) = 13(x + 1.1) = 13x + 14.3 J

According to the conservation law of energy, this potential energy is converted to spring elastic energy once it's compressed

E_p = E_k = kx^2/2 = 13x + 14.3

where k = 315 is the spring constant and x is the compressed length

315x^2 = 26x + 28.6

315x^2 - 26x - 28.6 = 0

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \frac{26 \pm \sqrt{26^2 - 4*(-28.6)*315}}{2*315}

x = \frac{26 \pm 191.6}{630}

x = 0.345 m or x = -0.263 m

Since x can only be positive we will pick the 0.345m

6 0
3 years ago
- Una carga de -5x10-6 C está situada a 20cm delante de otra
saul85 [17]

Answer:

F = 5.625\,N (Fuerza de repulsión/Repulsive force)

Explanation:

La fuerza ejercida entre las dos partículas se calcula por la Ley de Coulomb (The force exerted between the two particles is determined by Coulomb's Law):

F = k \cdot \frac{q_{1}\cdot q_{2}}{r^{2}}

Donde (Where):

k - Constante electrostática, medido en \frac{N\cdot m^{2}}{C^{2}} (Electrostatic constant, measured in \frac{N\cdot m^{2}}{C^{2}}).

q_{1}, q_{2} - Magnitudes de las cargas de cada partícula, medidos en Coulombs. (Magnitudes of charges from each particle, measured in Coulombs).

r - Distancia entre las partículas, medida en metros.

La fuerza electrostática es (Electrostatic force is):

F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \frac{(5\times 10^{-6}\,C)\cdot (5\times 10^{-6}\,C)}{(0.20\,m)^{2}}

F = 5.625\,N

Dado que las partículas tienen el mismo signo de carga, la fuerza es de repulsión. (Given that both particles have the same charge sign, the force is of repulsive nature)

3 0
2 years ago
The force of attraction between a star of mass M and a planet of mass m (where т «М) is: ЗЫ? a F where is the angular momentum o
Margaret [11]

Answer:

2b2t

Explanation:

2b2t

5 0
3 years ago
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