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4 years ago

Will Give Brainliest! 50 PTS :)

Physics

2 answers:

lakkis [162]4 years ago

8 0

-- Running at (14 km/hr), it takes the faster runner (8 km)/(14 km/hr) = 0.571 hr to run the complete race.

-- In that time, the slower runner covers (12 km/hr) x (0.571 hr) = 6.857 km .

-- At that point, he still has (8 km - 6.857 km) = 1.143 km or 1143 meters left to go.

lys-0071 [83]4 years ago

7 0

The slower runner is farther from the finish line. 2.06 km

8 - 5.94 =2.06

You might be interested in

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

3 years ago

A rotating wheel requires 5 s to rotate 38 revolutions. Its angular velocity at end of time interval 5-s is 79 rad/s. What is th

Lera25 [3.4K]

"Rotating wheel" is meant to indcate that the wheel is already rotating at the start. Denote the initial angular velocity by ω₀. Then the angular displacement θ at time t is

θ = ω₀t + 1/2 αt ²

while the angular velocity ω is

ω = ω₀ + αt

It takes 5 s for the wheel to rotate 38 times, or turn a total of (2π rad/rev) (38 rev) = 76π rad, as well as to reach an angular velocity of 79 rad/s, so that

76π rad = ω₀ (5 s) + 1/2 α (5 s)²

79 rad/s = ω₀ + α (5 s)

Solve the second equation for ω₀ and substitute into the first equation, then solve for α :

ω₀ = 79 rad/s - α (5 s)

==> 76π rad = (79 rad/s - α (5 s)) (5 s) + 1/2 α (5 s)²

==> 76π rad = (79 rad/s) (5 s) - 1/2 α (5 s)²

==> 1/2 α (5 s)² = (79 rad/s) (5 s) - 76π rad

==> α = ((79 rad/s) (5 s) - 76π rad) / (1/2 (5 s)²)

==> α ≈ 12.5 rad/s²

3 0

3 years ago

a body is projected at an angle of 30 degrees to the horizontal with a velocity of 15m/s, calculate the time it takes to reach t

andreev551 [17]

Explanation:

The vertical component the velocity of the projectile is 15 m/s x sin 30 = 7.5 m/s.

The body is accelarating downwards at 10 m/s^2.

This means that every second its upward velocity reduces by 10 m/s.

So if the body is travelling upwards at 7.5 m/s then how long does it take for the velocity to become 0?

(7.5 m/s) / (10 m/s^2) = 0.75 s

3 0

3 years ago

When one side of the Earth is at high tide, then the opposite side of the Earth is also at high tide.

Rudiy27

It is true because the near side of the moon is attracted by the gravitational force due to the alignment of the planet in one Straight line.The opposite side is due to the movement of earth towards moon side during alignment.Soon both side, there will be high tide at the same time.

8 0

4 years ago

A plane is flying at an elevation of 32000 feet. It is within sight of the airport and the pilot finds that the angle of depress

prohojiy [21]

Answer:x=79,202.77 ft

Explanation:

Given

plane is at a height of 32000 ft

angle of depression is $22^{\circ}$