The acceleration of gravity on the

What is the net power needed to change the speed of a 1600-kg sport utility vehicle from 15.0 m/s to 40.0 m/s in 4.00 seconds

Sergio039 [100]

Answer:

The net power needed to change the speed of the vehicle is 275,000 W

Explanation:

Given;

mass of the sport vehicle, m = 1600 kg

initial velocity of the vehicle, u = 15 m/s

final velocity of the vehicle, v = 40 m/s

time of motion, t = 4 s

The force needed to change the speed of the sport vehicle;

$F = \frac{m(v-u)}{t} \\\\F = \frac{1600(40-15)}{4} \\\\F = 10,000 \ N$

The net power needed to change the speed of the vehicle is calculated as;

$P_{net} = \frac{1}{2} F[u + v]\\\\P_{net} = \frac{1}{2} \times 10,000[15 + 40]\\\\P_{net} = 275,000 \ W$

3 0

2 years ago

Explain why an aton is neutral

kodGreya [7K]

Atoms ere electrically neutral because they have equal number of protons and electrons. If an atom lose or gain one or more electrons it becomes an ion.

4 0

3 years ago

If a crow flies west for 60 km and then south for 45 km, what is the direction of its displacement?

son4ous [18]

That's 105 km that he flew, or 65.2 miles ! I'm absolutely positive
that the crow must have landed and gotten some rest when you
weren't looking. But that had no effect on his displacement when
he got where he was going, so we can continue to solve the problem:

The displacement is the distance and direction from the place
where the crow took off to the place where he landed.

-- It's distance is the hypotenuse of the right triangle whose legs
are 60 km and 45 km.

D² = (60 km)² + (45 km)²

=    3,600 km² + 2,025 km² = 5,625 km²

   D = √(5625 km²) = 75 km .

-- It's direction is the angle whose tangent is (45 S / 60 W).

   tan⁻¹ (45/60) = tan⁻¹ (0.75) = 36.9° south of west

   = 53.1° west of south.

   = not exactly southwest but close.

7 0

3 years ago

The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be

bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

$F=\frac{Gm_{1} m_{2} }{d^{2} }$ ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

$1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }$

$d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}$

$d=\sqrt{1.48\times10^{17}}$

d = 3.85 x 10⁸ m

3 0

3 years ago

In Example 2.7, we investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on sol

igor_vitrenko [27]

Answer:

a) 20 seconds

b) No.

Explanation:

t = Time taken for jet to stop

u = Initial velocity = 100 m/s (given in the question)

v = Final velocity = 0 (because the jet will stop at the end)

s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)

a = Acceleration = -5.00 m/s² (slowing down, so it is negative)

a) Equation of motion

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s$

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=100\times 20+\frac{1}{2}\times -5\times 20^2\\\Rightarrow s=1000\ m$

The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.

5 0

3 years ago