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3 years ago

How are balanced & unbalanced forces related to net force?

Physics

1 answer:

Levart [38]3 years ago

4 0

Answer:

An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. ... A net force = unbalanced force. If however, the forces are balanced (in equilibrium) and there is no net force, the object will not accelerate and the velocity will remain constant.

Explanation:

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In a 5.00 km race, one runner runs at a steady 11.4 km/h and another runs at 14.7 km/h . How long does the faster runner have to

topjm [15]

Answer:

0.0986 h or 5 minutes 55 seconds.

Explanation:

Speed: This can be defined as the rate of change of distance of a body. The S.I unit of speed is m/s. Speed is a scalar quantity, because it can only be represented by magnitude alone.

Mathematically,

Speed = distance/time.

S = d/t ........................... Equation 1

making t the subject of the equation

t = d/S ......................... Equation 2

Form the question,

Time taken for the faster runner to reach the finish line

t₁ = d/S₁................... Equation 3

Where t₁ = time taken for the faster runner to reach the finish line, d = distance, S₁ = speed of the faster runner.

Given: d = 5.0 km, S₁ = 14.7 km/h.

Substituting into equation 3

t₁ = 5/14.7

t₁ = 0.340 h

Also,

t₂ = d/S₂................... Equation 4

Where t₂ = time taken for the slower runner to reached the finished line, d = distance, S₂ = speed of the slower runner.

Given: d = 5 km, S₂ = 11.4 km/h.

Substitute into equation 4,

t₂ = 5/11.4

t₂ = 0.4386 h.

The time the faster runner have to wait at the finish line to see the slower runner cross = t₂ - t₁ = 0.4386-0.340

The time the faster runner have to wait at the finish line to see the slower runner cross = 0.0986 h = 5 mins 55 s.

8 0

4 years ago

The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his

Alex17521 [72]

Answer:

22m/s

Explanation:

Mass, m=60 kg

Force constant, k=1300N/m

Restoring force, Fx=6500 N

Average friction force, f=50 N

Length of barrel, l=5m

y=2.5 m

Initial velocity, u=0

$F_x=kx$

Substitute the values

$6500=1300x$

$x=\frac{6500}{1300}=5$ m

Work done due to friction force

$W_f=fscos\theta$

We have $\theta=180^{\circ}$

Substitute the values

$W_f=50\times 5cos180^{\circ}$

$W_f=-250J$

Initial kinetic energy, Ki=0

Initial gravitational energy, $U_{grav,1}=0$ \

Initial elastic potential energy

$U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)$

$U_{el,1}=16250J$

Final elastic energy, $U_{el,2}=0$

Final kinetic energy, $K_f=\frac{1}{2}(60)v^2=30v^2$

Final gravitational energy, $U_{grav,2}=mgh=60\times 9.8\times 2.5$

Final gravitational energy, $U_{grav,2}=1470J$

Using work-energy theorem

$K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}$

Substitute the values

$0+0+16250-250=30v^2+1470+0$

$16000-1470=30v^2$

$14530=30v^2$

$v^2=\frac{14530}{30}$

$v=\sqrt{\frac{14530}{30}}$

$v=22m/s$

5 0

3 years ago

The displacement at any given time of an object is x = 6 sin 98t , where the symbols have their usual meanings. i). Proof that t

Fofino [41]

A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation, which are represented in the given function as 6 units and 98 rad/s respectively.

<h3>General wave equation for simple harmonic motion</h3>

y = A sinωt

where;

A is amplitude of the motion
ω is angular frequency

<h3>Amplitude of the oscillation</h3>

A = 6 units

<h3>Angular frequency of the wave</h3>

ω = 98 rad/s

A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation. Thus, the wave is executing simple harmonic motion.

Learn more about simple harmonic motion here: brainly.com/question/17315536

#SPJ1

5 0

2 years ago

An object stays at rest until what happens to it?

bixtya [17]

D. an outside or unbalanced force acts upon the object.

4 0

3 years ago

A(n) 28.3 g bullet is shot into a(n) 5004 g wooden block standing on a frictionless surface. The block, with the bullet in it, a

Natali [406]

Answer:

the speed of the bullet before striking the block is 302.3 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 28.3 g = 0.0283 kg

mass of the wooden block, m₂ = 5004 g = 5.004 kg

initial velocity of the block, u₂ = 0

final velocity of the bullet-wood system, v = 1.7 m/s

let the initial velocity of the bullet before striking the block = u₁

Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

0.0283u₁ + 5.004 x 0 = 1.7(0.0283 + 5.004)

0.0283u₁ = 8.5549

u₁ = 8.5549 / 0.0283

u₁ = 302.3 m/s

Therefore, the speed of the bullet before striking the block is 302.3 m/s.

4 0

3 years ago